problem 167 & 170 from leetcode;
https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/
https://leetcode.com/problems/two-sum-iii-data-structure-design/
先来看简单的167：给定一个排好序的数组，以及一个值，找出数组中相加等于该值的两个数，假设这样的值始终存在；
这个只要依次遍历该数组，假设当前值为x，那么只需要在数组中找到value - x；如果存在，直接返回，如果不存在，检查下一个值；因为数组是sorted，所以第二步查找只需要log N的时间；最坏情况是 n * log N;
public int[] twoSum(int[] numbers, int target) {
int[] result = new int[2];
for(int i = 1; i <= numbers.length; i++) {
int x = numbers[i - 1];
int y = target - x;
int j = Arrays.binarySearch(numbers, i, numbers.length, y);
if(j >= i) {
result[0] = i;
result[1] = j + 1;
break;
}
}
return result;
}题目170要求设计一种数据结构，支持add和find操作，下面是一个例子：
add(1); add(3); add(5);
find(4) -> true
find(7) -> false一开始我得想法是使用167的解法，add的时候，把输入加入到一个数组中，并且排好序；那么就可以直接使用167的find；但在输入很长的时候，很多的add和find，会TLE；简单的分析一下这种方式，（我使用的是insertion sort), 每次add排序，要O(n), 那么总的时间就是O(n * n);
后来我用AVL树，因为AVL树支持logN的insert/find操作；正好适合这个问题；以下是最终AC的代码：
public class TwoSum {
List<Integer> numbers = new ArrayList<>();
AVL avl = new AVL();
public void add(int number) {
avl.add(number);
numbers.add(number);
}
public boolean find(int value) {
for (int x : numbers) {
int y = value - x;
Node node = avl.find(y);
if (node == null) {
continue;
}
//when these two numbers equal, need to make sure there at least two numbers added;
if (x == y && node.count == 1) {
continue;
}
return true;
}
return false;
}
public static void main(String[] args) {
TwoSum twoSum = new TwoSum();
twoSum.add(1);
twoSum.add(2);
System.out.println(twoSum.find(1));
}
class AVL {
Node root;
private int height(Node root) {
if (root == null) {
return -1;
} else {
return root.height;
}
}
private Node insert(Node root, int value) {
if (root == null) {
root = new Node(value);
} else if (root.value == value) {
root.count += 1;
} else if (root.value < value) {
//go right;
root.right = insert(root.right, value);
if (height(root.right) - height(root.left) == 2) {
if (value > root.right.value) {
root = singleRotateWithRight(root);
} else {
root = doubleRotateWithRight(root);
}
}
} else {
//go left;
root.left = insert(root.left, value);
if (height(root.left) - height(root.right) == 2) {
if (value < root.left.value) {
root = singleRotateWithLeft(root);
} else {
root = doubleRotateWithLeft(root);
}
}
}
root.height = Math.max(height(root.left), height(root.right)) + 1;
return root;
}
private Node doubleRotateWithRight(Node k3) {
k3.right = singleRotateWithLeft(k3.right);
return singleRotateWithRight(k3);
}
private Node singleRotateWithRight(Node k2) {
Node k1 = k2.right;
k2.right = k1.left;
k1.left = k2;
k2.height = Math.max(height(k2.left), height(k2.right)) + 1;
k1.height = Math.max(height(k1.left), height(k1.right)) + 1;
return k1;
}
private Node doubleRotateWithLeft(Node k3) {
k3.left = singleRotateWithRight(k3.left);
return singleRotateWithLeft(k3);
}
private Node singleRotateWithLeft(Node k2) {
Node k1 = k2.left;
k2.left = k1.right;
k1.right = k2;
k2.height = Math.max(height(k2.left), height(k2.right)) + 1;
k1.height = Math.max(height(k1.left), height(k1.right)) + 1;
return k1;
}
public void add(int value) {
root = insert(root, value);
}
private Node find(Node root, int value) {
if (root == null) {
return null;
}
if (root.value == value && root.count == 0) {
return null;
}
if (root.value == value) {
return root;
}
if (value > root.value) {
return find(root.right, value);
} else {
return find(root.left, value);
}
}
public Node find(int value) {
return find(root, value);
}
public Node getRoot() {
return root;
}
}
static class Node {
final int value;
int count, height;
Node left, right;
Node(int value) {
this.value = value;
count = 1;
height = 0;
}
}
}果然这种方法非常的有效，似乎是Java里面最快的实现了；
BTW, AVL的实现是从教科书里面找出来的，直接去写，真没有这样的本事；
来源：oschina
链接：https://my.oschina.net/u/922297/blog/493164