leetcode 448. Find All Numbers Disappeared in an Array

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]

解法：
先遍历一遍数组，将碰到的数字对应的位置nums[i]上的数字变为负
再遍历一次数组，位置i上为正的数没有在数组中出现过，将i+1添加到结果中

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) 
    {
        int len = nums.size();
        for(int i = 0;i<len;i++)
        {
            int m = abs(nums[i])-1;
            nums[m] = nums[m]>0? -nums[m]:nums[m];
        }

        vector<int> res;
        for(int i = 0;i<len;i++)
        {
            if(nums[i] >0) 
                res.push_back(i+1);
        }
        return res;
    }
};