15算法课程 268. Missing Number



Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

Example 1

Input: [3,0,1]Output: 2

Example 2

Input: [9,6,4,2,3,5,7,0,1]Output: 8

solution:

现将序列元素排序，然后一次遍历i 从0 到n−1 ，若nums[i]!=i ，则i 便是缺失元素，若循环正常结束，则缺失元素为n ；

code:

class Solution {public:    int missingNumber(vector<int>& nums) {        if (nums.empty())            return 0;        sort(nums.begin(), nums.end());        for (unsigned i = 0; i < nums.size(); ++i)        {            if (nums[i] != i)                return i;        }        return nums.size();    }};