目录：
1、【剑指Offer学习】【面试题01：实现赋值运算符函数】
2、【剑指Offer学习】【面试题02：实现Singleton 模式——七种实现方式】
6、【剑指Offer学习】【面试题06：重建二叉树】
7、【剑指Offer学习】【面试题07：用两个栈实现队列】
8、【剑指Offer学习】【面试题08：旋转数组的最小数字】
10、【剑指Offer学习】【面试题10：二进制中1 的个数】
1、【剑指Offer学习】【面试题01：实现赋值运算符函数】
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题1：赋值运算符函数
// 题目：如下为类型CMyString的声明，请为该类型添加赋值运算符函数。
#include<cstring>
#include<cstdio>
class CMyString
{
public:
CMyString(char* pData = nullptr);
CMyString(const CMyString& str);
~CMyString(void);
CMyString& operator = (const CMyString& str);
void Print();
private:
char* m_pData;
};
CMyString::CMyString(char *pData)
{
if(pData == nullptr)
{
m_pData = new char[1];
m_pData[0] = '\0';
}
else
{
int length = strlen(pData);
m_pData = new char[length + 1];
strcpy(m_pData, pData);
}
}
CMyString::CMyString(const CMyString &str)
{
int length = strlen(str.m_pData);
m_pData = new char[length + 1];
strcpy(m_pData, str.m_pData);
}
CMyString::~CMyString()
{
delete[] m_pData;
}
CMyString& CMyString::operator = (const CMyString& str)
{
if(this == &str)
return *this;
delete []m_pData;
m_pData = nullptr;
m_pData = new char[strlen(str.m_pData) + 1];
strcpy(m_pData, str.m_pData);
return *this;
}
// ====================测试代码====================
void CMyString::Print()
{
printf("%s", m_pData);
}
void Test1()
{
printf("Test1 begins:\n");
char* text = "Hello world";
CMyString str1(text);
CMyString str2;
str2 = str1;
printf("The expected result is: %s.\n", text);
printf("The actual result is: ");
str2.Print();
printf(".\n");
}
// 赋值给自己
void Test2()
{
printf("Test2 begins:\n");
char* text = "Hello world";
CMyString str1(text);
str1 = str1;
printf("The expected result is: %s.\n", text);
printf("The actual result is: ");
str1.Print();
printf(".\n");
}
// 连续赋值
void Test3()
{
printf("Test3 begins:\n");
char* text = "Hello world";
CMyString str1(text);
CMyString str2, str3;
str3 = str2 = str1;
printf("The expected result is: %s.\n", text);
printf("The actual result is: ");
str2.Print();
printf(".\n");
printf("The expected result is: %s.\n", text);
printf("The actual result is: ");
str3.Print();
printf(".\n");
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
return 0;
}
2、【剑指Offer学习】【面试题02：实现Singleton 模式——七种实现方式】
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题2：实现Singleton模式
// 题目：设计一个类，我们只能生成该类的一个实例。
using System;
namespace _02_Singleton
{
public sealed class Singleton1
{
private Singleton1()
{
}
private static Singleton1 instance = null;
public static Singleton1 Instance
{
get
{
if (instance == null)
instance = new Singleton1();
return instance;
}
}
}
public sealed class Singleton2
{
private Singleton2()
{
}
private static readonly object syncObj = new object();
private static Singleton2 instance = null;
public static Singleton2 Instance
{
get
{
lock (syncObj)
{
if (instance == null)
instance = new Singleton2();
}
return instance;
}
}
}
public sealed class Singleton3
{
private Singleton3()
{
}
private static object syncObj = new object();
private static Singleton3 instance = null;
public static Singleton3 Instance
{
get
{
if (instance == null)
{
lock (syncObj)
{
if (instance == null)
instance = new Singleton3();
}
}
return instance;
}
}
}
public sealed class Singleton4
{
private Singleton4()
{
Console.WriteLine("An instance of Singleton4 is created.");
}
public static void Print()
{
Console.WriteLine("Singleton4 Print");
}
private static Singleton4 instance = new Singleton4();
public static Singleton4 Instance
{
get
{
return instance;
}
}
}
public sealed class Singleton5
{
Singleton5()
{
Console.WriteLine("An instance of Singleton5 is created.");
}
public static void Print()
{
Console.WriteLine("Singleton5 Print");
}
public static Singleton5 Instance
{
get
{
return Nested.instance;
}
}
class Nested
{
static Nested()
{
}
internal static readonly Singleton5 instance = new Singleton5();
}
}
class Program
{
static void Main(string[] args)
{
// 也会打印An instance of Singleton4 is created.
Singleton4.Print();
// 不会打印An instance of Singleton5 is created.
Singleton5.Print();
}
}
}
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题3（一）：找出数组中重复的数字
// 题目：在一个长度为n的数组里的所有数字都在0到n-1的范围内。数组中某些数字是重复的，但不知道有几个数字重复了，
// 也不知道每个数字重复了几次。请找出数组中任意一个重复的数字。例如，如果输入长度为7的数组{2, 3, 1, 0, 2, 5, 3}，
// 那么对应的输出是重复的数字2或者3。
#include <cstdio>
// 参数:
// numbers: 一个整数数组
// length: 数组的长度
// duplication: (输出) 数组中的一个重复的数字
// 返回值:
// true - 输入有效，并且数组中存在重复的数字
// false - 输入无效，或者数组中没有重复的数字
bool duplicate(int numbers[], int length, int* duplication)
{
if(numbers == nullptr || length <= 0)
return false;
for(int i = 0; i < length; ++i)
{
if(numbers[i] < 0 || numbers[i] > length - 1)
return false;
}
for(int i = 0; i < length; ++i)
{
while(numbers[i] != i)
{
if(numbers[i] == numbers[numbers[i]])
{
*duplication = numbers[i];
return true;
}
// 交换numbers[i]和numbers[numbers[i]]
int temp = numbers[i];
numbers[i] = numbers[temp];
numbers[temp] = temp;
}
}
return false;
}
// ====================测试代码====================
bool contains(int array[], int length, int number)
{
for(int i = 0; i < length; ++i)
{
if(array[i] == number)
return true;
}
return false;
}
void test(char* testName, int numbers[], int lengthNumbers, int expected[], int expectedExpected, bool validArgument)
{
printf("%s begins: ", testName);
int duplication;
bool validInput = duplicate(numbers, lengthNumbers, &duplication);
if(validArgument == validInput)
{
if(validArgument)
{
if(contains(expected, expectedExpected, duplication))
printf("Passed.\n");
else
printf("FAILED.\n");
}
else
printf("Passed.\n");
}
else
printf("FAILED.\n");
}
// 重复的数字是数组中最小的数字
void test1()
{
int numbers[] = { 2, 1, 3, 1, 4 };
int duplications[] = { 1 };
test("Test1", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int), true);
}
// 重复的数字是数组中最大的数字
void test2()
{
int numbers[] = { 2, 4, 3, 1, 4 };
int duplications[] = { 4 };
test("Test2", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int), true);
}
// 数组中存在多个重复的数字
void test3()
{
int numbers[] = { 2, 4, 2, 1, 4 };
int duplications[] = { 2, 4 };
test("Test3", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int), true);
}
// 没有重复的数字
void test4()
{
int numbers[] = { 2, 1, 3, 0, 4 };
int duplications[] = { -1 }; // not in use in the test function
test("Test4", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int), false);
}
// 没有重复的数字
void test5()
{
int numbers[] = { 2, 1, 3, 5, 4 };
int duplications[] = { -1 }; // not in use in the test function
test("Test5", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int), false);
}
// 无效的输入
void test6()
{
int* numbers = nullptr;
int duplications[] = { -1 }; // not in use in the test function
test("Test6", numbers, 0, duplications, sizeof(duplications) / sizeof(int), false);
}
void main()
{
test1();
test2();
test3();
test4();
test5();
test6();
while(1);
}
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题3（二）：不修改数组找出重复的数字
// 题目：在一个长度为n+1的数组里的所有数字都在1到n的范围内，所以数组中至
// 少有一个数字是重复的。请找出数组中任意一个重复的数字，但不能修改输入的
// 数组。例如，如果输入长度为8的数组{2, 3, 5, 4, 3, 2, 6, 7}，那么对应的
// 输出是重复的数字2或者3。
#include <iostream>
int countRange(const int* numbers, int length, int start, int end);
// 参数:
// numbers: 一个整数数组
// length: 数组的长度
// 返回值:
// 正数 - 输入有效，并且数组中存在重复的数字，返回值为重复的数字
// 负数 - 输入无效，或者数组中没有重复的数字
int getDuplication(const int* numbers, int length)
{
if(numbers == nullptr || length <= 0)
return -1;
int start = 1;
int end = length - 1;
while(end >= start)
{
int middle = ((end - start) >> 1) + start;
int count = countRange(numbers, length, start, middle);
if(end == start)
{
if(count > 1)
return start;
else
break;
}
if(count > (middle - start + 1))
end = middle;
else
start = middle + 1;
}
return -1;
}
int countRange(const int* numbers, int length, int start, int end)
{
if(numbers == nullptr)
return 0;
int count = 0;
for(int i = 0; i < length; i++)
if(numbers[i] >= start && numbers[i] <= end)
++count;
return count;
}
// ====================测试代码====================
void test(const char* testName, int* numbers, int length, int* duplications, int dupLength)
{
int result = getDuplication(numbers, length);
for(int i = 0; i < dupLength; ++i)
{
if(result == duplications[i])
{
std::cout << testName << " passed." << std::endl;
return;
}
}
std::cout << testName << " FAILED." << std::endl;
}
// 多个重复的数字
void test1()
{
int numbers[] = { 2, 3, 5, 4, 3, 2, 6, 7 };
int duplications[] = { 2, 3 };
test("test1", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));
}
// 一个重复的数字
void test2()
{
int numbers[] = { 3, 2, 1, 4, 4, 5, 6, 7 };
int duplications[] = { 4 };
test("test2", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));
}
// 重复的数字是数组中最小的数字
void test3()
{
int numbers[] = { 1, 2, 3, 4, 5, 6, 7, 1, 8 };
int duplications[] = { 1 };
test("test3", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));
}
// 重复的数字是数组中最大的数字
void test4()
{
int numbers[] = { 1, 7, 3, 4, 5, 6, 8, 2, 8 };
int duplications[] = { 8 };
test("test4", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));
}
// 数组中只有两个数字
void test5()
{
int numbers[] = { 1, 1 };
int duplications[] = { 1 };
test("test5", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));
}
// 重复的数字位于数组当中
void test6()
{
int numbers[] = { 3, 2, 1, 3, 4, 5, 6, 7 };
int duplications[] = { 3 };
test("test6", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));
}
// 多个重复的数字
void test7()
{
int numbers[] = { 1, 2, 2, 6, 4, 5, 6 };
int duplications[] = { 2, 6 };
test("test7", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));
}
// 一个数字重复三次
void test8()
{
int numbers[] = { 1, 2, 2, 6, 4, 5, 2 };
int duplications[] = { 2 };
test("test8", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));
}
// 没有重复的数字
void test9()
{
int numbers[] = { 1, 2, 6, 4, 5, 3 };
int duplications[] = { -1 };
test("test9", numbers, sizeof(numbers) / sizeof(int), duplications, sizeof(duplications) / sizeof(int));
}
// 无效的输入
void test10()
{
int* numbers = nullptr;
int duplications[] = { -1 };
test("test10", numbers, 0, duplications, sizeof(duplications) / sizeof(int));
}
void main()
{
test1();
test2();
test3();
test4();
test5();
test6();
test7();
test8();
test9();
test10();
}
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题4：二维数组中的查找
// 题目：在一个二维数组中，每一行都按照从左到右递增的顺序排序，每一列都按
// 照从上到下递增的顺序排序。请完成一个函数，输入这样的一个二维数组和一个
// 整数，判断数组中是否含有该整数。
#include <cstdio>
bool Find(int* matrix, int rows, int columns, int number)
{
bool found = false;
if(matrix != nullptr && rows > 0 && columns > 0)
{
int row = 0;
int column = columns - 1;
while(row < rows && column >=0)
{
if(matrix[row * columns + column] == number)
{
found = true;
break;
}
else if(matrix[row * columns + column] > number)
-- column;
else
++ row;
}
}
return found;
}
// ====================测试代码====================
void Test(char* testName, int* matrix, int rows, int columns, int number, bool expected)
{
if(testName != nullptr)
printf("%s begins: ", testName);
bool result = Find(matrix, rows, columns, number);
if(result == expected)
printf("Passed.\n");
else
printf("Failed.\n");
}
// 1 2 8 9
// 2 4 9 12
// 4 7 10 13
// 6 8 11 15
// 要查找的数在数组中
void Test1()
{
int matrix[][4] = {{1, 2, 8, 9}, {2, 4, 9, 12}, {4, 7, 10, 13}, {6, 8, 11, 15}};
Test("Test1", (int*)matrix, 4, 4, 7, true);
}
// 1 2 8 9
// 2 4 9 12
// 4 7 10 13
// 6 8 11 15
// 要查找的数不在数组中
void Test2()
{
int matrix[][4] = {{1, 2, 8, 9}, {2, 4, 9, 12}, {4, 7, 10, 13}, {6, 8, 11, 15}};
Test("Test2", (int*)matrix, 4, 4, 5, false);
}
// 1 2 8 9
// 2 4 9 12
// 4 7 10 13
// 6 8 11 15
// 要查找的数是数组中最小的数字
void Test3()
{
int matrix[][4] = {{1, 2, 8, 9}, {2, 4, 9, 12}, {4, 7, 10, 13}, {6, 8, 11, 15}};
Test("Test3", (int*)matrix, 4, 4, 1, true);
}
// 1 2 8 9
// 2 4 9 12
// 4 7 10 13
// 6 8 11 15
// 要查找的数是数组中最大的数字
void Test4()
{
int matrix[][4] = {{1, 2, 8, 9}, {2, 4, 9, 12}, {4, 7, 10, 13}, {6, 8, 11, 15}};
Test("Test4", (int*)matrix, 4, 4, 15, true);
}
// 1 2 8 9
// 2 4 9 12
// 4 7 10 13
// 6 8 11 15
// 要查找的数比数组中最小的数字还小
void Test5()
{
int matrix[][4] = {{1, 2, 8, 9}, {2, 4, 9, 12}, {4, 7, 10, 13}, {6, 8, 11, 15}};
Test("Test5", (int*)matrix, 4, 4, 0, false);
}
// 1 2 8 9
// 2 4 9 12
// 4 7 10 13
// 6 8 11 15
// 要查找的数比数组中最大的数字还大
void Test6()
{
int matrix[][4] = {{1, 2, 8, 9}, {2, 4, 9, 12}, {4, 7, 10, 13}, {6, 8, 11, 15}};
Test("Test6", (int*)matrix, 4, 4, 16, false);
}
// 鲁棒性测试，输入空指针
void Test7()
{
Test("Test7", nullptr, 0, 0, 16, false);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
Test7();
return 0;
}
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题5：替换空格
// 题目：请实现一个函数，把字符串中的每个空格替换成"%20"。例如输入“We are happy.”，
// 则输出“We%20are%20happy.”。
#include <cstdio>
#include <cstring>
/*length 为字符数组str的总容量，大于或等于字符串str的实际长度*/
void ReplaceBlank(char str[], int length)
{
if(str == nullptr && length <= 0)
return;
/*originalLength 为字符串str的实际长度*/
int originalLength = 0;
int numberOfBlank = 0;
int i = 0;
while(str[i] != '\0')
{
++ originalLength;
if(str[i] == ' ')
++ numberOfBlank;
++ i;
}
/*newLength 为把空格替换成'%20'之后的长度*/
int newLength = originalLength + numberOfBlank * 2;
if(newLength > length)
return;
int indexOfOriginal = originalLength;
int indexOfNew = newLength;
while(indexOfOriginal >= 0 && indexOfNew > indexOfOriginal)
{
if(str[indexOfOriginal] == ' ')
{
str[indexOfNew --] = '0';
str[indexOfNew --] = '2';
str[indexOfNew --] = '%';
}
else
{
str[indexOfNew --] = str[indexOfOriginal];
}
-- indexOfOriginal;
}
}
// ====================测试代码====================
void Test(char* testName, char str[], int length, char expected[])
{
if(testName != nullptr)
printf("%s begins: ", testName);
ReplaceBlank(str, length);
if(expected == nullptr && str == nullptr)
printf("passed.\n");
else if(expected == nullptr && str != nullptr)
printf("failed.\n");
else if(strcmp(str, expected) == 0)
printf("passed.\n");
else
printf("failed.\n");
}
// 空格在句子中间
void Test1()
{
const int length = 100;
char str[length] = "hello world";
Test("Test1", str, length, "hello%20world");
}
// 空格在句子开头
void Test2()
{
const int length = 100;
char str[length] = " helloworld";
Test("Test2", str, length, "%20helloworld");
}
// 空格在句子末尾
void Test3()
{
const int length = 100;
char str[length] = "helloworld ";
Test("Test3", str, length, "helloworld%20");
}
// 连续有两个空格
void Test4()
{
const int length = 100;
char str[length] = "hello world";
Test("Test4", str, length, "hello%20%20world");
}
// 传入nullptr
void Test5()
{
Test("Test5", nullptr, 0, nullptr);
}
// 传入内容为空的字符串
void Test6()
{
const int length = 100;
char str[length] = "";
Test("Test6", str, length, "");
}
//传入内容为一个空格的字符串
void Test7()
{
const int length = 100;
char str[length] = " ";
Test("Test7", str, length, "%20");
}
// 传入的字符串没有空格
void Test8()
{
const int length = 100;
char str[length] = "helloworld";
Test("Test8", str, length, "helloworld");
}
// 传入的字符串全是空格
void Test9()
{
const int length = 100;
char str[length] = " ";
Test("Test9", str, length, "%20%20%20");
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
Test7();
Test8();
Test9();
return 0;
}
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题6：从尾到头打印链表
// 题目：输入一个链表的头结点，从尾到头反过来打印出每个结点的值。
#include "..\Utilities\List.h"
#include <stack>
void PrintListReversingly_Iteratively(ListNode* pHead)
{
std::stack<ListNode*> nodes;
ListNode* pNode = pHead;
while(pNode != nullptr)
{
nodes.push(pNode);
pNode = pNode->m_pNext;
}
while(!nodes.empty())
{
pNode = nodes.top();
printf("%d\t", pNode->m_nValue);
nodes.pop();
}
}
void PrintListReversingly_Recursively(ListNode* pHead)
{
if(pHead != nullptr)
{
if (pHead->m_pNext != nullptr)
{
PrintListReversingly_Recursively(pHead->m_pNext);
}
printf("%d\t", pHead->m_nValue);
}
}
// ====================测试代码====================
void Test(ListNode* pHead)
{
PrintList(pHead);
PrintListReversingly_Iteratively(pHead);
printf("\n");
PrintListReversingly_Recursively(pHead);
}
// 1->2->3->4->5
void Test1()
{
printf("\nTest1 begins.\n");
ListNode* pNode1 = CreateListNode(1);
ListNode* pNode2 = CreateListNode(2);
ListNode* pNode3 = CreateListNode(3);
ListNode* pNode4 = CreateListNode(4);
ListNode* pNode5 = CreateListNode(5);
ConnectListNodes(pNode1, pNode2);
ConnectListNodes(pNode2, pNode3);
ConnectListNodes(pNode3, pNode4);
ConnectListNodes(pNode4, pNode5);
Test(pNode1);
DestroyList(pNode1);
}
// 只有一个结点的链表: 1
void Test2()
{
printf("\nTest2 begins.\n");
ListNode* pNode1 = CreateListNode(1);
Test(pNode1);
DestroyList(pNode1);
}
// 空链表
void Test3()
{
printf("\nTest3 begins.\n");
Test(nullptr);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
return 0;
}
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题7：重建二叉树
// 题目：输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输
// 入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,
// 2, 4, 7, 3, 5, 6, 8}和中序遍历序列{4, 7, 2, 1, 5, 3, 8, 6}，则重建出
// 图2.6所示的二叉树并输出它的头结点。
#include "..\Utilities\BinaryTree.h"
#include <exception>
#include <cstdio>
BinaryTreeNode* ConstructCore(int* startPreorder, int* endPreorder, int* startInorder, int* endInorder);
BinaryTreeNode* Construct(int* preorder, int* inorder, int length)
{
if(preorder == nullptr || inorder == nullptr || length <= 0)
return nullptr;
return ConstructCore(preorder, preorder + length - 1,
inorder, inorder + length - 1);
}
BinaryTreeNode* ConstructCore
(
int* startPreorder, int* endPreorder,
int* startInorder, int* endInorder
)
{
// 前序遍历序列的第一个数字是根结点的值
int rootValue = startPreorder[0];
BinaryTreeNode* root = new BinaryTreeNode();
root->m_nValue = rootValue;
root->m_pLeft = root->m_pRight = nullptr;
if(startPreorder == endPreorder)
{
if(startInorder == endInorder && *startPreorder == *startInorder)
return root;
else
throw std::exception("Invalid input.");
}
// 在中序遍历中找到根结点的值
int* rootInorder = startInorder;
while(rootInorder <= endInorder && *rootInorder != rootValue)
++ rootInorder;
if(rootInorder == endInorder && *rootInorder != rootValue)
throw std::exception("Invalid input.");
int leftLength = rootInorder - startInorder;
int* leftPreorderEnd = startPreorder + leftLength;
if(leftLength > 0)
{
// 构建左子树
root->m_pLeft = ConstructCore(startPreorder + 1, leftPreorderEnd,
startInorder, rootInorder - 1);
}
if(leftLength < endPreorder - startPreorder)
{
// 构建右子树
root->m_pRight = ConstructCore(leftPreorderEnd + 1, endPreorder,
rootInorder + 1, endInorder);
}
return root;
}
// ====================测试代码====================
void Test(char* testName, int* preorder, int* inorder, int length)
{
if(testName != nullptr)
printf("%s begins:\n", testName);
printf("The preorder sequence is: ");
for(int i = 0; i < length; ++ i)
printf("%d ", preorder[i]);
printf("\n");
printf("The inorder sequence is: ");
for(int i = 0; i < length; ++ i)
printf("%d ", inorder[i]);
printf("\n");
try
{
BinaryTreeNode* root = Construct(preorder, inorder, length);
PrintTree(root);
DestroyTree(root);
}
catch(std::exception& exception)
{
printf("Invalid Input.\n");
}
}
// 普通二叉树
// 1
// / \
// 2 3
// / / \
// 4 5 6
// \ /
// 7 8
void Test1()
{
const int length = 8;
int preorder[length] = {1, 2, 4, 7, 3, 5, 6, 8};
int inorder[length] = {4, 7, 2, 1, 5, 3, 8, 6};
Test("Test1", preorder, inorder, length);
}
// 所有结点都没有右子结点
// 1
// /
// 2
// /
// 3
// /
// 4
// /
// 5
void Test2()
{
const int length = 5;
int preorder[length] = {1, 2, 3, 4, 5};
int inorder[length] = {5, 4, 3, 2, 1};
Test("Test2", preorder, inorder, length);
}
// 所有结点都没有左子结点
// 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
void Test3()
{
const int length = 5;
int preorder[length] = {1, 2, 3, 4, 5};
int inorder[length] = {1, 2, 3, 4, 5};
Test("Test3", preorder, inorder, length);
}
// 树中只有一个结点
void Test4()
{
const int length = 1;
int preorder[length] = {1};
int inorder[length] = {1};
Test("Test4", preorder, inorder, length);
}
// 完全二叉树
// 1
// / \
// 2 3
// / \ / \
// 4 5 6 7
void Test5()
{
const int length = 7;
int preorder[length] = {1, 2, 4, 5, 3, 6, 7};
int inorder[length] = {4, 2, 5, 1, 6, 3, 7};
Test("Test5", preorder, inorder, length);
}
// 输入空指针
void Test6()
{
Test("Test6", nullptr, nullptr, 0);
}
// 输入的两个序列不匹配
void Test7()
{
const int length = 7;
int preorder[length] = {1, 2, 4, 5, 3, 6, 7};
int inorder[length] = {4, 2, 8, 1, 6, 3, 7};
Test("Test7: for unmatched input", preorder, inorder, length);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
Test7();
return 0;
}
8、【剑指Offer学习】【面试题08：旋转数组的最小数字】
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题8：二叉树的下一个结点
// 题目：给定一棵二叉树和其中的一个结点，如何找出中序遍历顺序的下一个结点？
// 树中的结点除了有两个分别指向左右子结点的指针以外，还有一个指向父结点的指针。
#include <stdio.h>
struct BinaryTreeNode
{
int m_nValue;
BinaryTreeNode* m_pLeft;
BinaryTreeNode* m_pRight;
BinaryTreeNode* m_pParent;
};
BinaryTreeNode* GetNext(BinaryTreeNode* pNode)
{
if(pNode == nullptr)
return nullptr;
BinaryTreeNode* pNext = nullptr;
if(pNode->m_pRight != nullptr)
{
BinaryTreeNode* pRight = pNode->m_pRight;
while(pRight->m_pLeft != nullptr)
pRight = pRight->m_pLeft;
pNext = pRight;
}
else if(pNode->m_pParent != nullptr)
{
BinaryTreeNode* pCurrent = pNode;
BinaryTreeNode* pParent = pNode->m_pParent;
while(pParent != nullptr && pCurrent == pParent->m_pRight)
{
pCurrent = pParent;
pParent = pParent->m_pParent;
}
pNext = pParent;
}
return pNext;
}
// ==================== 辅助代码用来构建二叉树 ====================
BinaryTreeNode* CreateBinaryTreeNode(int value)
{
BinaryTreeNode* pNode = new BinaryTreeNode();
pNode->m_nValue = value;
pNode->m_pLeft = nullptr;
pNode->m_pRight = nullptr;
pNode->m_pParent = nullptr;
return pNode;
}
void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, BinaryTreeNode* pRight)
{
if(pParent != nullptr)
{
pParent->m_pLeft = pLeft;
pParent->m_pRight = pRight;
if(pLeft != nullptr)
pLeft->m_pParent = pParent;
if(pRight != nullptr)
pRight->m_pParent = pParent;
}
}
void PrintTreeNode(BinaryTreeNode* pNode)
{
if(pNode != nullptr)
{
printf("value of this node is: %d\n", pNode->m_nValue);
if(pNode->m_pLeft != nullptr)
printf("value of its left child is: %d.\n", pNode->m_pLeft->m_nValue);
else
printf("left child is null.\n");
if(pNode->m_pRight != nullptr)
printf("value of its right child is: %d.\n", pNode->m_pRight->m_nValue);
else
printf("right child is null.\n");
}
else
{
printf("this node is null.\n");
}
printf("\n");
}
void PrintTree(BinaryTreeNode* pRoot)
{
PrintTreeNode(pRoot);
if(pRoot != nullptr)
{
if(pRoot->m_pLeft != nullptr)
PrintTree(pRoot->m_pLeft);
if(pRoot->m_pRight != nullptr)
PrintTree(pRoot->m_pRight);
}
}
void DestroyTree(BinaryTreeNode* pRoot)
{
if(pRoot != nullptr)
{
BinaryTreeNode* pLeft = pRoot->m_pLeft;
BinaryTreeNode* pRight = pRoot->m_pRight;
delete pRoot;
pRoot = nullptr;
DestroyTree(pLeft);
DestroyTree(pRight);
}
}
// ====================测试代码====================
void Test(char* testName, BinaryTreeNode* pNode, BinaryTreeNode* expected)
{
if(testName != nullptr)
printf("%s begins: ", testName);
BinaryTreeNode* pNext = GetNext(pNode);
if(pNext == expected)
printf("Passed.\n");
else
printf("FAILED.\n");
}
// 8
// 6 10
// 5 7 9 11
void Test1_7()
{
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode9 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNode11 = CreateBinaryTreeNode(11);
ConnectTreeNodes(pNode8, pNode6, pNode10);
ConnectTreeNodes(pNode6, pNode5, pNode7);
ConnectTreeNodes(pNode10, pNode9, pNode11);
Test("Test1", pNode8, pNode9);
Test("Test2", pNode6, pNode7);
Test("Test3", pNode10, pNode11);
Test("Test4", pNode5, pNode6);
Test("Test5", pNode7, pNode8);
Test("Test6", pNode9, pNode10);
Test("Test7", pNode11, nullptr);
DestroyTree(pNode8);
}
// 5
// 4
// 3
// 2
void Test8_11()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNode5, pNode4, nullptr);
ConnectTreeNodes(pNode4, pNode3, nullptr);
ConnectTreeNodes(pNode3, pNode2, nullptr);
Test("Test8", pNode5, nullptr);
Test("Test9", pNode4, pNode5);
Test("Test10", pNode3, pNode4);
Test("Test11", pNode2, pNode3);
DestroyTree(pNode5);
}
// 2
// 3
// 4
// 5
void Test12_15()
{
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode2, nullptr, pNode3);
ConnectTreeNodes(pNode3, nullptr, pNode4);
ConnectTreeNodes(pNode4, nullptr, pNode5);
Test("Test12", pNode5, nullptr);
Test("Test13", pNode4, pNode5);
Test("Test14", pNode3, pNode4);
Test("Test15", pNode2, pNode3);
DestroyTree(pNode2);
}
void Test16()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
Test("Test16", pNode5, nullptr);
DestroyTree(pNode5);
}
int main(int argc, char* argv[])
{
Test1_7();
Test8_11();
Test12_15();
Test16();
}
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题9：用两个栈实现队列
// 题目：用两个栈实现一个队列。队列的声明如下，请实现它的两个函数appendTail
// 和deleteHead，分别完成在队列尾部插入结点和在队列头部删除结点的功能。
#include "Queue.h"
// ====================测试代码====================
void Test(char actual, char expected)
{
if(actual == expected)
printf("Test passed.\n");
else
printf("Test failed.\n");
}
int main(int argc, char* argv[])
{
CQueue<char> queue;
queue.appendTail('a');
queue.appendTail('b');
queue.appendTail('c');
char head = queue.deleteHead();
Test(head, 'a');
head = queue.deleteHead();
Test(head, 'b');
queue.appendTail('d');
head = queue.deleteHead();
Test(head, 'c');
queue.appendTail('e');
head = queue.deleteHead();
Test(head, 'd');
head = queue.deleteHead();
Test(head, 'e');
return 0;
}
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题9：用两个栈实现队列
// 题目：用两个栈实现一个队列。队列的声明如下，请实现它的两个函数appendTail
// 和deleteHead，分别完成在队列尾部插入结点和在队列头部删除结点的功能。
#pragma once
#include <stack>
#include <exception>
using namespace std;
template <typename T> class CQueue
{
public:
CQueue(void);
~CQueue(void);
// 在队列末尾添加一个结点
void appendTail(const T& node);
// 删除队列的头结点
T deleteHead();
private:
stack<T> stack1;
stack<T> stack2;
};
template <typename T> CQueue<T>::CQueue(void)
{
}
template <typename T> CQueue<T>::~CQueue(void)
{
}
template<typename T> void CQueue<T>::appendTail(const T& element)
{
stack1.push(element);
}
template<typename T> T CQueue<T>::deleteHead()
{
if(stack2.size()<= 0)
{
while(stack1.size()>0)
{
T& data = stack1.top();
stack1.pop();
stack2.push(data);
}
}
if(stack2.size() == 0)
throw new exception("queue is empty");
T head = stack2.top();
stack2.pop();
return head;
}
10、【剑指Offer学习】【面试题10：二进制中1 的个数】
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题10：斐波那契数列
// 题目：写一个函数，输入n，求斐波那契（Fibonacci）数列的第n项。
#include <cstdio>
// ====================方法1：递归====================
long long Fibonacci_Solution1(unsigned int n)
{
if(n <= 0)
return 0;
if(n == 1)
return 1;
return Fibonacci_Solution1(n - 1) + Fibonacci_Solution1(n - 2);
}
// ====================方法2：循环====================
long long Fibonacci_Solution2(unsigned n)
{
int result[2] = {0, 1};
if(n < 2)
return result[n];
long long fibNMinusOne = 1;
long long fibNMinusTwo = 0;
long long fibN = 0;
for(unsigned int i = 2; i <= n; ++ i)
{
fibN = fibNMinusOne + fibNMinusTwo;
fibNMinusTwo = fibNMinusOne;
fibNMinusOne = fibN;
}
return fibN;
}
// ====================方法3：基于矩阵乘法====================
#include <cassert>
struct Matrix2By2
{
Matrix2By2
(
long long m00 = 0,
long long m01 = 0,
long long m10 = 0,
long long m11 = 0
)
:m_00(m00), m_01(m01), m_10(m10), m_11(m11)
{
}
long long m_00;
long long m_01;
long long m_10;
long long m_11;
};
Matrix2By2 MatrixMultiply
(
const Matrix2By2& matrix1,
const Matrix2By2& matrix2
)
{
return Matrix2By2(
matrix1.m_00 * matrix2.m_00 + matrix1.m_01 * matrix2.m_10,
matrix1.m_00 * matrix2.m_01 + matrix1.m_01 * matrix2.m_11,
matrix1.m_10 * matrix2.m_00 + matrix1.m_11 * matrix2.m_10,
matrix1.m_10 * matrix2.m_01 + matrix1.m_11 * matrix2.m_11);
}
Matrix2By2 MatrixPower(unsigned int n)
{
assert(n > 0);
Matrix2By2 matrix;
if(n == 1)
{
matrix = Matrix2By2(1, 1, 1, 0);
}
else if(n % 2 == 0)
{
matrix = MatrixPower(n / 2);
matrix = MatrixMultiply(matrix, matrix);
}
else if(n % 2 == 1)
{
matrix = MatrixPower((n - 1) / 2);
matrix = MatrixMultiply(matrix, matrix);
matrix = MatrixMultiply(matrix, Matrix2By2(1, 1, 1, 0));
}
return matrix;
}
long long Fibonacci_Solution3(unsigned int n)
{
int result[2] = {0, 1};
if(n < 2)
return result[n];
Matrix2By2 PowerNMinus2 = MatrixPower(n - 1);
return PowerNMinus2.m_00;
}
// ====================测试代码====================
void Test(int n, int expected)
{
if(Fibonacci_Solution1(n) == expected)
printf("Test for %d in solution1 passed.\n", n);
else
printf("Test for %d in solution1 failed.\n", n);
if(Fibonacci_Solution2(n) == expected)
printf("Test for %d in solution2 passed.\n", n);
else
printf("Test for %d in solution2 failed.\n", n);
if(Fibonacci_Solution3(n) == expected)
printf("Test for %d in solution3 passed.\n", n);
else
printf("Test for %d in solution3 failed.\n", n);
}
int main(int argc, char* argv[])
{
Test(0, 0);
Test(1, 1);
Test(2, 1);
Test(3, 2);
Test(4, 3);
Test(5, 5);
Test(6, 8);
Test(7, 13);
Test(8, 21);
Test(9, 34);
Test(10, 55);
Test(40, 102334155);
return 0;
}
来源：https://www.cnblogs.com/agui125/p/12020184.html