[leetcode]448. Find All Numbers Disappeared in an Array

Analysis

pre做完一身轻松~—— [嘻嘻~]

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.
给出一个数组，数组大小为n，本来数组中应该是从1到n的n个数，但是有些书重复出现了，导致其他数没有出现，判断数组中有哪些数没有出现过。把出现过的数的相应位置的数变为其相反数，然后遍历数组，看那些位置上数字仍然大于0，则为缺失的数。

Implement

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        vector<int> res;
        for(auto num:nums){
            int t = abs(num) - 1;
            nums[t] = abs(nums[t]) * -1;
        }
        for(int i=0; i<nums.size(); i++){
            if(nums[i] > 0)
                res.push_back(i+1);
        }
        return res;
    }
};