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Problem
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]
Solution
如果有一个数字i出现两次，那么nums[i-1]的符号就是正的。
class Solution {
public:vector<int> findDisappearedNumbers(vector<int>& nums) {for (auto x: nums) {x = abs(x);if (nums[x - 1] > 0) nums[x - 1] *= -1;}vector<int> res;for (int i = 0; i < nums.size(); i ++ )if (nums[i] > 0)res.push_back(i + 1);return res;}
};
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