【LeetCode】268. Missing Number

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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

 

等差数列求和，易得缺失值。

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

转载于:https://www.cnblogs.com/ganganloveu/p/4758875.html