LeetCode Single Number I & II 都符合两个问题额外要求的 通用解法 与 思考过程

Single Number

Given an array of integers, every element appears twice except for one. Find that single one.Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Single Number II

Given an array of integers, every element appears three times except for one. Find that single one.Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?.1. 既然可以在查找循环外添加循环，那么添加什么循环好呢？排序吧，没错！排序的数列干什么都方便！（感觉这个结论也适用很多地方, 虽然计算复杂度不能达到O(n)）2. 利用map处理，效率也接近O(n)，不过空间复杂度也要O（n）3. 利用bit process，位运算下面是Single number I 的位运算解法，思路就是每位bit出现2次就清零，所以可以不断异或运算得出最终结果:

int singleNumber(int A[], int n){int ans = 0;for (int i = 0; i < n; i++) ans ^= A[i];return ans;}

Single number II 的位运算两种解法-- 参考leetcode论坛的代码：方法一，int的32个bit逐个处理-这个方法还算简单的了：

  int singleNumber(int A[], int n) {int  result =0;for(int i = 0; i< 32; i++){int c = 0;int bit = 1<<i;for(int j = 0; j<n; j++){if (A[j] & bit)c++;}if (c%3)result |=bit;}return result;}

这个方法比较普遍，对于任何的n times 都试用。其实异或的方法和其思想是一指的。