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?

题目大意：给定一个数组，长度为n，包含从0～n中的n个数，问哪个数不存在？

思路：加减就可以了。

    public int missingNumber(int[] nums) {

        if(nums==null||nums.length==0){

            return 0;

        }

        int res = 0;

        for(int i=0;i<nums.length;i++){

            res+=nums[i];

            res-=i+1;

        }

        return -res;

    }