leetcode 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?

思路：

　　既然给出的数字是连续的，并且只有一个数字缺失，那么我们只需要求出前n项的和（n*(n+1)/2）再减去所给数字的和就是缺失的那个数字了。

Code：

int missingNumber(int* nums, int numsSize) {
    int sum1=0, sum2=(numsSize * (numsSize+1)) / 2;
    for(int i=0; i<numsSize; i++)
    {
        sum1 += nums[i];
    }
    return sum2 - sum1;
}