位運算（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?

排序後再找時間複雜度不合題意。

0……n異或，而後nums[i]異或：

 1 public class Solution {
 2     public int missingNumber(int[] nums) {
 3         int n = nums.length;
 4         int res = 0;
 5         for(int i=1; i<=n; i++) {
 6             res = res ^ i;
 7         }
 8         for(int i=0; i<n; i++) {
 9              res = res ^ nums[i];
10         }
11         return res;
12     }
13 }

用異或解決的題目還有 「有幾個數，其中只有一個數有奇數個，其他的數有偶數個，找出奇數個的那個數」