LeetCode OJ：Missing Number (丢失的数)

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?

题目实际上很简单，就是要求求出一段o-n中缺了一个数的数列中找出缺失的那个，我一开始的想法是这样的：

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int sz = nums.size();
        for (int i = 0; i < sz - 1; ++i){
            if (nums[i] != nums[i + 1])
                return nums[i + 1];
        }
        return nums[sz - 1] + 1;
    }
};

代码很简单，就是遍历比较而已。但是很明显的，不太符合题目对于常数项空间复杂度的要求，出去翻了翻答案，别人家的小孩是这样写的：

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int sz = nums.size();
        int total = (sz + 1)*sz / 2;
        for (int i = 0; i < sz; ++i){
            total -= nums[i];
        }
        return total;
    }
};

这种被吊打的感觉，有点像高斯当时算随手算出5050吊打同伴小孩的情况。

java代码如下：

public class Solution {
    public int missingNumber(int[] nums) {
        int sum = nums.len * (nums.len + 1)/2; //length大小是n-1
        for(int i = 0; i < nums.length; ++i){
            sum -= nums[i];
        }
        return sum;
    }
}