349. 两个数组的交集

输入: nums1 = [1,2,2,1], nums2 = [2,2]输出: [2]

输入: nums1 = [4,9,5], nums2 = [9,4,9,8,4]输出: [9,4]

/* * @lc app=leetcode id=349 lang=javascript * * [349] Intersection of Two Arrays * * https://leetcode.com/problems/intersection-of-two-arrays/description/ * * algorithms * Easy (53.11%) * Total Accepted:    203.6K * Total Submissions: 380.9K * Testcase Example:  '[1,2,2,1]\n[2,2]' * * Given two arrays, write a function to compute their intersection. * * 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] * * * Note: * * * Each element in the result must be unique. * The result can be in any order. * * * * *//** * @param {number[]} nums1 * @param {number[]} nums2 * @return {number[]} */var intersection = function(nums1, nums2) {    const visited = {};    const ret = [];    for(let i = 0; i < nums1.length; i++) {        const num = nums1[i];        visited[num] = num;    }    for(let i = 0; i < nums2.length; i++) {        const num = nums2[i];        if (visited[num] !== undefined) {            ret.push(num);            visited[num] = undefined;        }    }    return ret;};

class Solution:    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:        visited, result = {}, []        for num in nums1:            visited[num] = num        for num in nums2:            if num in visited:                result.append(num)                visited.pop(num)        return result    # 另一种解法：利用 Python 中的集合进行计算    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:        return set(nums1) & set(nums2)