1.描述
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?
2.分析
3.代碼
class Solution {
public:
int missingNumber(vector<int>& nums) {
int n = nums.size();
return (1 + n) * n / 2 - accumulate(nums.begin(), nums.end(), 0);
}
};
