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?

给定一个数字包含数字从 0..n，其中缺了一个数字，找到缺失的数字。

如果不考虑复杂度的话，可以先求0...n的数字的总和，然后减去给定数组的和，差值即为缺失的数字。

位操作中有特性 0^k = k, k^k=0 所以可以把数组中所有数字都xor一遍，然后和0...n中所有数字xor，最后得到的结果就是缺失的数

public int missingNumber(int[] nums) {
   int result = 0;
   for (int i : nums) {
       result^=i;
   }
   for (int i=0;i<=nums.length;i++) {
       result^=i;
   }
   return result;
}