这两道题都是求两个数组之间的重复元素,因此把它们放在一起. 原题地址: 349 Intersection of Two Arrays :https://leetcode.com/problems/intersection-of-two-arrays/description/ 350 Intersection of Two Arrays II:https://leetcode.com/problems/intersection-of-two-arrays-ii/description/ 题目&&am…

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…

Given two arrays, write a function to compute their intersection. Example:Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2]. Note: Each element in the result must be unique. The result can be in any order. 题目标签:Hash Table 题目给了我们两个 array, 让我们找到在两…

Given two arrays, write a function to compute their intersection. Example:Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2]. Note: Each element in the result must be unique. The result can be in any order. Subscribe to see which companies asked…

Given two arrays, write a function to compute their intersection. Example: Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2]. Note: Each element in the result must be unique. The result can be in any order. 思路:利用 C++ 中的set 容器.…

题目要求 Given two arrays, write a function to compute their intersection. 题目分析及思路 给定两个数组,要求得到它们之中共同拥有的元素列表(列表中元素是唯一的).可以将所给数组转成集合,利用集合的交集的概念,最后将结果转成列表即可. python代码 class Solution: def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]: n…

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] 思路 代码 class Solution { public int[] intersection(int…

Given two arrays, write a function to compute their intersection. Example 1: Input: nums1 = [1,2,2,1], nums2 = [2,2] Output: [2,2] Example 2: Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4] Output: [4,9] Note: Each element in the result should appear as…

283. Move Zeroes var moveZeroes = function(nums) { var num1=0,num2=1; while(num1!=num2){ nums.forEach(function(x,y){ if(x===0){ nums.splice(y,1); nums.push(0); } num1 = nums ; }); nums.forEach(function(x,y){ if(x===0){ nums.splice(y,1); nums.push(0);…

最近,在用解决LeetCode问题的时候,做了349: Intersection of Two Arrays这个问题,就是求两个列表的交集.我这种弱鸡,第一种想法是把问题解决,而不是分析复杂度,于是写出了如下代码: class Solution(object): def intersection(self, nums1, nums2): """ :type nums1: List[int] :type nums2: List[int] :rtype: List[int] &q…