LeetCode 448. Find All Numbers Disappeared in an Array （数组查缺）

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Example:

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

输入一个长度为n的数组（元素有重复），查找[1, n]之间缺少的数。

思路：用unique()函数把原数组去重，遍历其中缺少的数字。

    vector<int> findDisappearedNumbers(vector<int>& nums) {        int len = nums.size();        vector<int> nums2=nums;        vector<int> ans(0);        sort(nums.begin(),nums.end());        nums.erase(unique(nums.begin(),nums.end()),nums.end());        for(int i=0;i<len;i++)        {            if(nums[i]!=i+1)            {                nums.insert(nums.begin()+i,i+1);                ans.push_back(i+1);            }                        }        return ans;    }