268. 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?
时间复杂度O(n)比较好办，但是空间复杂度O(1)难一些。
将所有的i与nums[i]异或，得到的结果就是少掉的那个数。

public class Solution {
public int missingNumber(int[] nums) {
int xor=0,i;
for(i=0;i<nums.length;i++){
xor^=i^nums[i];
}
return xor^i;
}
}另一种方法为求和：sum(i)-sum(nums[i]). 但是结果很可能会溢出。