442 Find all duplicates in an array_442. find all duplicates in an array-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_38396940/article/details/124658927

这篇博客介绍了两种解决在给定长度为n的整数数组中找出所有出现两次的整数的方法。第一种方法利用位操作，通过将正负号用来标记元素出现次数，当遇到一个元素时，将其对应的位取反，第二次遇到时就会发现其已变为负数。第二种方法使用哈希表，遍历数组时将元素作为键，如果已存在则将该元素添加到结果中，否则将其值设为1。两种方法都在O(n)时间内完成并只使用了常数额外空间。示例输入为[4,3,2,7,8,2,3,1]，输出为[2,3]。

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The description of problem

Given an integer array nums of length n where all the integers of nums are in the range [1, n] 
and each integer appears once or twice, return an array of all the integers that appears twice.

You must write an algorithm that runs in O(n) time and uses only constant extra space.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-all-duplicates-in-an-array

an example

Input: nums = [4,3,2,7,8,2,3,1]
Output: [2,3]

Solution 1

intuition

use positive and negative signs to indicate the appearance times for a specific element.
specifically, when you just found a elements nums[i], we could use minus sign add to the nums[nums[i] - 1]
easily to see, when you just found the second elements in your array, you will do the above operation once again.

The codes

#include <vector>
#include <iostream>
#include <cmath>
using namespace std;
class Solution {
public:
    vector<int> findDuplicates(vector<int>& nums) {
		vector<int> res;
		for (auto num : nums) {
			int flag_pos = abs(num) - 1;
			if (nums[flag_pos] < 0) {
				res.push_back(abs(num));
			} else {
				nums[flag_pos] = -nums[flag_pos];
			}
		}
		return res;
    }
};
int main() 
{
	Solution s;
	vector<int> nums = {4,3,2,7,8,2,3,1};
	vector<int> res = s.findDuplicates(nums);
	for (int i = 0; i < res.size(); i++) {
		cout << res[i] << " ";
	}
	cout << endl;
}

The corresponding results

2 3

The solution 2

The intuition

use the data structure of map to find the elements appearing twice in the array

The codes

#include <vector>
#include <iostream>
#include <unordered_map>
using namespace std;
class Solution {
public:
    vector<int> findDuplicates(vector<int>& nums) {
		vector<int> res;
		// initialize a hash table with 0
		unordered_map<int, int> hash;
		for (auto num : nums) {
			if (hash[num] == 1) {
				res.emplace_back(num);
			} else {
				hash[num] = 1;
			}
		}
		return res;
		
    }
};
int main() 
{
	Solution s;
	vector<int> nums = {4,3,2,7,8,2,3,1};
	vector<int> res = s.findDuplicates(nums);
	for (int i = 0; i < res.size(); i++) {
		cout << res[i] << " ";
	}
	cout << endl;
}