【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]...
一、前言 做了两题才慢慢摸清了leetcode的操作。 二、题349 Intersection of Two Arrays Given two arrays, write a function to compute their intersection. classSolution(object):defintersection(self, nums1, nums2):""":type nums1: List[int] :type nums2: List[int] :rtype: List[int]"""nums3...
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
Given two arrays, write a function to compute their intersection. Example: Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2]. Note: 1、Each element in the result must be unique. 2、The result can be in any order. 要完成的函数: vector<int> intersection(vector<int>& ...
349. Intersection of Two Arrays刷题笔记 考察集合的交运算。python的set底层是用哈希表实现的 class Solution: def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]: return list(set(nums1) & set(nums2)) 1. 2.
1// 349. Intersection of Two Arrays 2// https://leetcode.com/problems/intersection-of-two-arrays/description/ 3// 时间复杂度: O(nlogn) 4// 空间复杂度: O(n) 5class Solution { 6public: 7 vector<int> intersection(vector<int>& nums1, vector<int>& nums2) { 8 9 set<int> record(...
We firstly sort the two arrays and use two pointers to compare the elements in the two arrays from the beginning to the end. If the two elements are the same, forward both pointers and put the element to the return array If the two elements are not the same, move the smaller one's ...
Python as an example cw clockwise, ccw counter... def ring_area(x, y): """x and y as separate lists or arrays from references above - assumes first and last points are the same so uses slices """ a0 = np.dot(x[1:], y[:-1]) a1 = np.dot(x[:-1], y[1:]) ar...
I check the return code ofBrepBrep()and take affirmative action if False is returned. I didn't really need to do this, if the function failed then the crvs and pts arrays should be null, which I also check for, but this does provide more detailed information to the end user. There ...
If this condition is satisfied simultaneously for two segments then we know that they will cross at some point. Each factor of the 'C' arrays is essentially a matrix containing the numerators of the signed distances between points of one curve and line segments of the other....