然后又看到了350. Intersection of Two Arrays II,这次的结果是两个数组的交集,但是可以有重复元素了,要运行O(n)的话,这次直接想到了用空间换时间,无非是使用hash了,Python的字典就是hash实现的,于是写了: 1classSolution(object):2defintersect(self, nums1, nums2):3"""4:type nums1: List[int]5:type ...
【leetcode python】Intersection of Two Arrays #-*- coding: UTF-8 -*- #求两个集合的交集 class Solution(object): def intersection(self, nums1, nums2): resultList=list(set(nums1).intersection(set(nums2))) return resultList sol=Solution() print sol.intersection(nums1 =[1, 2, 2, 1]...
def intersection(self, nums1: [int], nums2: [int]) -> [int]: # python 内置集合运算 return list(set(nums1) & set(nums2)) 源代码文件在这里。 ©本文首发于 何睿的博客 ,欢迎转载,转载需保留
python 两个数组的交集 intersection of two arrays,给定两个数组,写一个函数来计算它们的交集。例子:给定num1=[1,2,2,1],nums2=[2,2],返回 [2].提示:每个在结果中的元素必定是唯一的。我们可以不考虑输出结果的顺序。classSolution(object):defintersection(self,nums1,n
每天一算:Intersection of Two Arrays II leetcode上第350号问题:Intersection of Two Arrays II 给定两个数组,编写一个函数来计算它们的交集。 示例 1: 输入: nums1 = [1,2,2,1], nums2 = [2,2] 输出: [2,2] 示例 2: 输入: nums1 = [4,9,5], nums2 = [9,4,9,8,4] 输出: [4,9]...
LeetCode 0349 - Intersection of Two Arrays Intersection of Two Arrays Desicription Given two arrays, write a function to compute their intersection. Example 1: 代码语言:javascript 复制 Input:nums1=[1,2,2,1],nums2=[2,2]Output:[2]
使用Collections模块的Counter类,目的是用来跟踪值出现的次数。Counter类是一个无序的容器类型,以字典的键值对形式存储,其中元素作为key,其计数作为value。计数值可以是任意的Interger(包括0和负数)。 代码如下: 解题思路3:直接使用collections模块中Counter类的算术和集合操作。 应用于题目的代码如下:...
https://leetcode.com/explore/interview/card/top-interview-questions-easy/92/array/674/leetcode.com/explore/interview/card/top-interview-questions-easy/92/array/674/ Solutions: 1. Sort and merge: We firstly sort the two arrays and use two pointers to compare the elements in the two arrays...
Prior to utilizing the code provided below, kindly employ the install lodash module through thenpm install lodashcommand. Opting forIntersection of two arrays. Javascript The following codes are included in this document:// Requiring the lodash libraryconst _ = require("lodash");// Original array...
P = InterX(L1x,L1y,L2x,L2y) returns the intersection points of two curves L1 and L2. The curves L1,L2 can be either closed or open. P = InterX(L1x,L1y) returns the self-intersection points of L1. To keep the code simple, the points at which the curve is tangent to itself ...