LeetCode 448.Find All Numbers Disappeared in an Array

Description:

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]

分析：

hash[i]用来标记i是否在nums数组中出现过，然后判断hash[i]的真假将没有出现过的放入res数组中，返回res数组

代码如下：

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
	vector<int> findDisappearedNumbers(vector<int>& nums) {
		vector<int> res;
		vector<bool> hash(nums.size() + 1, false);
		for (int i = 0; i < nums.size(); i++) {
			hash[nums[i]] = true;
		}
		for (int i = 1; i < hash.size(); i++) {
			if (hash[i] == false) {
				res.push_back(i);
			}
		}
		return res;
	}
};
int main() {
	Solution s;
	int n;
	cin >> n;
	vector<int> nums(n);
	for (int i = 0; i < n; i++) {
		cin >> nums[i];
	}
	vector<int> temp = s.findDisappearedNumbers(nums);
	for (int i = 0; i < temp.size(); i++) {
		cout << temp[i] << " ";
	}
	return 0;
}