Single Number II

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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?

题目解析：数组都是3个3个出现，找出唯一的数字，这个是在2个2个出现的升级版，2个2个直接异或即可

解题思路：先用快排将数组排序，然后维护一个times和result字段，另其初始为2，如果下一个数跟前一个数相等，则times--，当times为01说明3个已经过去，将times重新赋值为3，result=nums[i]，按照这种方式遍历一边即可，这里的这个小技巧来自剑指offer面试题29，下面上AC代码

 public int singleNumber(int[] array) {        // 安全性检查 数组是否符合3*n+1;float len = array.length;if(len==1.0)return array[0];if ((len - 1) % 3 != 0 || len < 4)return -1;quick_sort(array, 0, array.length - 1);int FindNum = array[0];int times = 2;for (int i = 1; i < len; i++) {if (times == 0) {// 换下一个数，times置为3重新开始FindNum = array[i];times = 3;}if (FindNum == array[i])times--;}return FindNum;    }    // 快速排序private static void quick_sort(int[] arr, int low, int high) {// 解决和合并if (low <= high) {int mid = partition(arr, low, high);// 递归quick_sort(arr, low, mid - 1);quick_sort(arr, mid + 1, high);}}private static int partition(int[] arr, int low, int high) {// 分解int pivot = arr[high];int i = low - 1;int temp;for (int j = low; j < high; j++) {if (arr[j] < pivot) {i++;temp = arr[i];arr[i] = arr[j];arr[j] = temp;}}// 交换中间元素和privottemp = arr[i + 1];arr[i + 1] = arr[high];arr[high] = temp;return i + 1;}

思路二：由于2个2个使用异或，即不进位的2进制加法，这里我们可以定义一个不进位的3进制加法也可解决

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