448. Find All Numbers Disappeared in an Array 难度：easy

题目：

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]

思路：

把原数组的数放在应该的位置，然后再重新遍历标志数组。

程序：

class Solution {public:    vector<int> findDisappearedNumbers(vector<int>& nums) {        vector<int> res;        for(int i = 0;i < nums.size();i++)        {            if(nums[i] != i + 1)            {                int j = nums[i];                while(nums[j - 1] != j)                {                    int temp = nums[j - 1];                    nums[j - 1] = j;                    j = temp;                }            }        }        for(int i = 0;i < nums.size();i++)            if(i + 1 != nums[i])                res.push_back(i + 1);                return res;    }};