位运算（4）——Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

排序后再找时间复杂度不合题意。

0……n异或，然后nums[i]异或：

 public class Solution {
     public int missingNumber(int[] nums) {
         int n = nums.length;
         int res = 0;
         for(int i=1; i<=n; i++) {
             res = res ^ i;
         }
         for(int i=0; i<n; i++) {
              res = res ^ nums[i];
         }
         return res;
     }
 }

用异或解决的题目还有 “有几个数，其中只有一个数有奇数个，其余的数有偶数个，找出奇数个的那个数”