Skip to content

Latest commit

 

History

History
224 lines (183 loc) · 4.45 KB

README_EN.md

File metadata and controls

224 lines (183 loc) · 4.45 KB
Raw
comments difficulty edit_url tags
true
Easy
Array
Hash Table
Two Pointers
Binary Search
Sorting

349. Intersection of Two Arrays

中文文档

Description

Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must be unique and you may return the result in any order.

 

Example 1:

Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2]

Example 2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [9,4]
Explanation: [4,9] is also accepted.

 

Constraints:

  • 1 <= nums1.length, nums2.length <= 1000
  • 0 <= nums1[i], nums2[i] <= 1000

Solutions

Solution 1

Python3

class Solution:
    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
        return list(set(nums1) & set(nums2))

Java

class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        boolean[] s = new boolean[1001];
        for (int x : nums1) {
            s[x] = true;
        }
        List<Integer> ans = new ArrayList<>();
        for (int x : nums2) {
            if (s[x]) {
                ans.add(x);
                s[x] = false;
            }
        }
        return ans.stream().mapToInt(Integer::intValue).toArray();
    }
}

C++

class Solution {
public:
    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
        bool s[1001];
        memset(s, false, sizeof(s));
        for (int x : nums1) {
            s[x] = true;
        }
        vector<int> ans;
        for (int x : nums2) {
            if (s[x]) {
                ans.push_back(x);
                s[x] = false;
            }
        }
        return ans;
    }
};

Go

func intersection(nums1 []int, nums2 []int) (ans []int) {
	s := [1001]bool{}
	for _, x := range nums1 {
		s[x] = true
	}
	for _, x := range nums2 {
		if s[x] {
			ans = append(ans, x)
			s[x] = false
		}
	}
	return
}

JavaScript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersection = function (nums1, nums2) {
    const s = Array(1001).fill(false);
    for (const x of nums1) {
        s[x] = true;
    }
    const ans = [];
    for (const x of nums2) {
        if (s[x]) {
            ans.push(x);
            s[x] = false;
        }
    }
    return ans;
};

C#

public class Solution {
    public int[] Intersection(int[] nums1, int[] nums2) {
        List<int> result = new List<int>();
        HashSet<int> arr1 = new(nums1);
        HashSet<int> arr2 = new(nums2);
        foreach (int x in arr1) {
            if (arr2.Contains(x)) {
                result.Add(x);
            }
        }
        return result.ToArray();
    }
}

PHP

class Solution {
    /**
     * @param Integer[] $nums1
     * @param Integer[] $nums2
     * @return Integer[]
     */
    function intersection($nums1, $nums2) {
        $rs = [];
        $set1 = array_values(array_unique($nums1));
        $set2 = array_values(array_unique($nums2));
        for ($i = 0; $i < count($set1); $i++) {
            $hashmap[$set1[$i]] = 1;
        }
        for ($j = 0; $j < count($set2); $j++) {
            if ($hashmap[$set2[$j]]) {
                array_push($rs, $set2[$j]);
            }
        }
        return $rs;
    }
}

Solution 2

JavaScript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersection = function (nums1, nums2) {
    return Array.from(new Set(nums1)).filter(num => new Set(nums2).has(num));
};