349. Intersection of Two Arrays

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

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

Java Code
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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++ Code
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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 Code
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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 Code
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/**
 * @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# Code
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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 Code
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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 Code
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/**
 * @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));
};