题目描述：

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:

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Input:
[4,3,2,7,8,2,3,1]

Output:
[5,6]

寻找没有出现过的数字，因为数组中的元素满足1 ≤ a[i] ≤ n，所以可以利用下标来保存一个数是否出现过，把一个数放到它对应的下标处，然后再找出哪些下标与元素不对应就是没有出现过的数。

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