题目描述: 第一次提交: class Solution: def containsDuplicate(self, nums: List[int]) -> bool: nums2 = set(nums) return not len(nums2)==len(nums) 方法二:哈希表 class Solution: def containsDuplicate(self, nums: List[int]) -> bool: v
easy vision: #-*- coding: utf-8 -*-import cursesfrom random import randrange, choice from collections import defaultdict #Defines valid inputs to not allow for input errors actions = ['Up', 'Left', 'Down', 'Right',
class Solution { public boolean canFinish(int numCourses, int[][] prerequisites) { int[] indegrees=new int[numCourses]; int[][] graph=new int[numCourses][numCourses]; for(int i=0;i<prerequisites.length;i++) {
class Solution {public: bool canFinish(int numCourses, vector<vector<int>>& prerequisites) { //验证是否为DAG,每次验证指向的是否已经存在于当前图中 //建图 vector<int> indegree(numCourses,0);//入度 vector<vector<int
题目如下: There are a total of n courses you have to take, labeled from 0 to n-1. Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1] Given the total number of courses and
题目描述 http://uoj.ac/problem/207 题解 因为这道题有删边和加边的操作,所以我们不能再链上操作,只能在点上操作。 考虑一些正确性玄学的算法。 我们给每一次链加随机一个权值,这样对于每次询问就查一下这条边分成的两块中的权值异或和是否等于当前所有链的权值异或和即可。 代码 #
知识补充: ①:ssprintf: int sprintf(char *str, const char *format, ...) 发送格式化输出到 str 所指向的字符串。 char str[80]; sprintf(str, "Pi 的值 = %f", M_PI); puts(str); ②:strchr 如果需要对字符串中的单个字符进行查找, 那