1，Vaild Palindrome
1 bool isPalindrome(string& s) {
2 transform(s.begin(), s.end(), s.begin(), tolower); // 把字符全部转换成小写
3 int left = 0;
4 int right = s.length()-1;
5 while (left < right) {
6 if (!isalnum(s[left])) ++left;
7 else if (!isalnum(s[right])) --right;
8 else if (s[left] != s[right]) return false;
9 else {
10 ++left;
11 --right;
12 }
13 }
14 return true;
15 }2，Implement strStr
1 int strStr(const string& haystack, const string& needle) { // KMP 算法，BM 算法
2 if (needle.empty()) return 0;
3 const int N = haystack.length() - needle.length();
4 for (int i = 0; i <= N; ++i) {
5 int j = i; // 源串索引
6 int k = 0; // 字串索引
7 while (haystack[j] == needle[k] && j < haystack.length() && k < needle.length()) {
8 ++j;
9 ++k;
10 }
11 if (k == needle.length()) return i;
12 }
13 return -1;
14 }3，String to Integer(atoi)
1 int myAtoi(const string& str) {
2 const int n = str.length();
3 int sign = 1;
4 int i = 0;
5 int num = 0;
6
7 while (str[i] == ' ' && i < n) ++i; // 前面的字符处理
8 if (str[i] == '+') {
9 sign = 1;
10 ++i;
11 }
12 else if (str[i] == '-') {
13 sign = -1;
14 ++i;
15 }
16 for (; i < n; ++i) {
17 if (str[i] < '0' || str[i] > '9')
18 break;
19 num = num * 10 + str[i] - '0';
20
21 if (sign == 1 && num > INT_MAX)
22 return INT_MAX;
23 else if (sign == -1 && num > INT_MIN + 1)
24 return INT_MIN;
25 }
26 return sign * num;
27 }4，Add Binary
1 string addBinary(string a, string b) { // 思路同 Add Binary(链表类型题) plusOne(数组类型题)
2 string result;
3 int carry = 0;
4 reverse(a.begin(), a.end()); // 先反转两个数组，从低位开始相加
5 reverse(b.begin(), b.end());
6 const int n = a.length() > b.length() ? a.length() : b.length();
7 for (int i = 0; i < n; ++i) {
8 const int ai = i < a.length() ? a[i]-'0' : 0;
9 const int bi = i < b.length() ? b[i]-'0' : 0;
10 int value = (ai + bi + carry) % 2;
11 carry = (ai + bi + carry) / 2;
12 result.insert(result.begin(), value + '0');
13 }
14 if (carry == 1)
15 result.insert(result.begin(), '1');
16 return result;
17 }5，Longest Palindromic Substring(未实现)
6，Regular Expression Matching
1 bool isMatchI(const char *s, const char *p) { //递归版 有挑战的一道题目
2 if (*p == '\0') return *s == '\0';
3
4 //next char is not '*',then match current character
5 if (*(p + 1) != '*') {
6 if (*p == *s || (*p == '.' && *s != '\0')) { // correctly match
7 return isMatchI(s + 1, p + 1);
8 }
9 else { // failed match
10 return false;
11 }
12 }
13 else { // next char is '*'
14 while (*p == *s || (*p == '.' && *s != '\0')) {
15 if (isMatchI(s, p + 2)) {
16 return true;
17 }
18 s++;
19 }
20 return isMatchI(s, p + 2);
21 }
22 }7，Wildcard Matching
1 bool isMatchII1(const char *s, const char *p) { // 递归版
2 if (*p == '*') {
3 while (*p == '*') ++p; // skip continuous '*'
4 if (*p == '\0') return true;
5 while (*s != '\0' && !isMatchII1(s, p)) ++s;
6
7 return *s != '\0';
8 }
9 else if (*p == '\0' || *s == '\0')
10 return *p == *s;
11 else if (*p == *s || *p == '?')
12 return isMatchII1(++s, ++p);
13 else
14 return false;
15 }
16 bool isMatchII2(const char *s, const char *p) { // 迭代版
17 bool star = false;
18 const char *str = s; // str 可以变，*str 不能变
19 const char *ptr = p;
20
21 while (*str != '\0') {
22 switch (*ptr) {
23 case '?':
24 str++;
25 ptr++;
26 break;
27 case '*':
28 star = true;
29 while (*ptr == '*') ++p;
30 if (*ptr == '\0') return true;
31 break;
32 default:
33 if (*str != *ptr) {
34 if (!star)
35 return false;
36 str++;
37 }
38 }
39 }
40 while (*ptr == '*') ++ptr;
41 return (*ptr == '\0');
42
43 }8，Longest Common Prefix
1 string longestCommonPrefix1(vector<string>& strs) { // 纵向扫描
2 if (strs.empty()) return "";
3
4 for (int index = 0; index < strs[0].size(); ++index) { // 选取第一个字符串和其它字符串进行比较
5 for (int i = 1; i < strs.size(); ++i) {
6 if (strs[i][index] != strs[0][index])
7 return strs[0].substr(0, index);
8 }
9 }
10 return strs[0];
11 }
12
13 string longestCommonPrefix2(vector<string>& strs) { // 横向扫描
14 if (strs.empty()) return "";
15
16 int right_most = strs[0].size() - 1;
17 for (size_t i = 1; i < strs.size(); ++i) {
18 for (int j = 0; j <= right_most; ++j) {
19 if (strs[i][j] != strs[0][j])
20 right_most = j - 1;
21 }
22 }
23 return strs[0].substr(0, right_most + 1);
24 }9，Valid Number
1 bool isNumber(const char *s) { //
2 char* endptr;
3 strtod(s, &endptr);
4
5 if (endptr == s)
6 return false;
7
8 for (; *endptr; ++endptr) {
9 if (!isspace(*endptr))
10 return false;
11 }
12 return true;
13 }10，Integet to Roman
1 string intToRoman1(int num) {
2 const int radix[] = { 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 };
3 const string symbol[] = { "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" };
4
5 string roman;
6 for (size_t i = 0; num > 0; ++i) {
7 int count = num / radix[i];
8 num %= radix[i];
9 for (; count > 0; --count)
10 roman += symbol[i];
11 }
12 return roman;
13 }
14
15 string intToRoman2(int num) {
16 string res = "";
17 vector<int> val = { 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 };
18 vector<string> str = { "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I" };
19 for (int i = 0; i < val.size(); ++i) {
20 while (num >= val[i]) {
21 num -= val[i];
22 res += str[i];
23 }
24 }
25 return res;
26 }11，Roman to Integer
1 int romanToInt(const string& s) {
2 int result = 0;
3 unordered_map<char, int> mapping;
4 mapping['I'] = 1;
5 mapping['V'] = 5;
6 mapping['X'] = 10;
7 mapping['L'] = 50;
8 mapping['C'] = 100;
9 mapping['D'] = 500;
10 mapping['M'] = 1000;
11
12 for (size_t i = 0; i < s.size(); i++) {
13 if (i > 0 && (mapping[i] > mapping[i - 1])) {
14 result += (mapping[s[i]] - 2 * mapping[s[i - 1]]);
15 }
16 else {
17 result += mapping[s[i]];
18 }
19 }
20 return result;
21 }12，Count and Say
1 string getNext(const string& s) {
2 stringstream ss;
3 for (auto i = s.begin(); i != s.end();) {
4 auto j = find_if(i, s.end(), bind1st(not_equal_to<char>(), *i)); // 需要看看函数怎么用
5 ss << distance(i, j) << *i;
6 i = j;
7 }
8 return ss.str();
9 }
10
11 string countAndSay(int n) {
12 string s("1");
13 while (--n) {
14 s = getNext(s);
15 }
16 return s;
17 }13，Anagrams(回文构词法)
1 vector<string> anagrams(vector<string>& strs) {
2 unordered_map<string, vector<string>> group;
3 for ( auto s = strs.begin(); s != strs.end();++s) {
4 string key = *s;
5 sort(key.begin(), key.end()); // 会修改原数据
6 group[key].push_back(*s);
7 }
8 vector<string> result;
9 for (auto it = group.cbegin(); it != group.cend(); ++it) {
10 if (it->second.size() > 1)
11 result.insert(result.end(), it->second.begin(), it->second.end());
12 }
13 return result;
14 }14，Simplify Path
1 string simplifyPath(const string& path) {
2 string dir;
3 stack<string> stk;
4 string result;
5
6 for (auto i = path.begin(); i != path.end();) {
7 ++i;
8 auto j = find(i, path.end(), '/');
9 dir = string(i, j); /* 获取两个 / / 之间的内容 */
10
11 if (!dir.empty() && dir !="/" && dir != ".") {
12 if (dir == "..") {
13 if (!stk.empty())
14 stk.pop();
15 }
16 else
17 stk.push(dir);
18 }
19 i = j;
20 }
21 //stringstream out; // 可以用字符串流保存，然后通过 out.str() 转化成字符串返回
22 if (stk.empty())
23 //out << "/";
24 result = "/";
25 else {
26 while (!stk.empty()) {
27 string s = stk.top();
28 stk.pop();
29 //out << '/' << s;
30 result += "/" + s;
31 }
32 }
33 // return out.str();
34 return result;
35 }15，Length of Last Word
1 int lengthOfLastWord1(const string& s) { // STL::find_if,find_if_not,distance
2 auto first = find_if(s.rbegin(), s.rend(), isalpha);
3 auto last = find_if_not(first, s.rend(), isalpha);
4 return distance(first, last);
5 }
6
7 int lengthOfLastWord2(const string& s) {
8 int len = 0;
9 for (int i = 0; i < s.length(); ++i) {
10 if (s[i] != ' ')
11 ++len;
12 else
13 len = 0;
14 }
15 return len;
16 }
c++常用函数
1, split
#include<vector>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
vector<string> split(string str, const char* c) {
char *cstr, *p;
vector<string> res;
cstr = new char[str.length()+1];
strcpy(cstr, str.c_str());
p = strtok(cstr, c);
while(p != NULL) {
res.push_back(p);
p = strtok(NULL, c);
}
delete cstr; // 释放内存
return res;
}
int main() {
string s = "a AND B";
const char *c = " ";
vector<string> ans;
ans = split(s, c);
for(int i=0;i<ans.size();i++) {
cout << ans[i] << endl;
}
return 0;
}
以上题目来源于：http://www.github.com/soulmachine/leetcode
来源：oschina
链接：https://my.oschina.net/u/4282181/blog/3630671