Chinese Postman Problem is a very famous hard problem in graph theory. The problem is to find a shortest closed path or circuit that visits every edge of a (connected) undirected graph. When the graph has an Eulerian Circuit (a closed walk that cover…
Description 题目描述 You are given an polynomial of x consisting of only addition marks, multiplication marks, brackets, single digit numbers, and of course the letter x. For example, a valid polynomial would be: (1+x)*(1+x*x+x+5)+1*x*x. You are required…
传送门 Description You have 4 cards each containing a number from 1 to 9. You need to judge whether they could operated through *, /, +, -, (, )to get the value of 24. Example 1: Input: [4, 1, 8, 7] Output: True Explanation: (8-4) * (7-1) = 24 Example 2…
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is: [ "((()))", "(()())", "(())()", "()(())", "()()()" ] 给定一…
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is: [ "((()))", "(()())", "(())()", "()(())", "()()()" ] 给出 …
题目: 给出 n 代表生成括号的对数,请你写出一个函数,使其能够生成所有可能的并且有效的括号组合. 例如,给出 n =3,生成结果为: [ "((()))", "(()())", "(())()", "()(())", "()()()" ]看到这个题,是合格的括号,有n个左括号就有n个右括号,那通过回溯法可以解决.代码如下: class Solution { public List<String>…
[抄题]: Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is: [ "((()))", "(()())", "(())()", "()(())", "()()()"…
把左右括号剩余的次数记录下来,传入回溯函数. 判断是否得到结果的条件就是剩余括号数是否都为零. 注意判断左括号是否剩余时,加上left>0的判断条件!否则会memory limited error! 判断右括号时要加上i==1的条件,否则会出现重复的答案. 同样要注意在回溯回来后ans.pop_back() class Solution { public: void backTrack(string ans, int left, int right, vector<string>&…
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is: "((()))", "(()())", "(())()", "()(())", "()()()" 求出所有可能的…