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: [ "((()))", "(()())", "(())()", "()(())", "()()()" ] 思路:…

一.题目说明 这个题目是22. Generate Parentheses,简单来说,输入一个数字n,输出n对匹配的小括号. 简单考虑了一下,n=0,输出"";n=1,输出"()":n=2,输出"()()","(())"... 二.我的解法 我考虑了一下,基本上2种思路,其1是暴力破解法,就是列出所有可能情况,然后判断哪些是匹配的. 汗颜的是,我没做出来.找到的答案: #include<iostream> #incl…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 22: Generate Parentheseshttps://oj.leetcode.com/problems/generate-parentheses/ Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.For exa…

22. Generate Parentheses . Generate Parentheses Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = , a solution set is: [ "((()))", "(()())", "(())()"…

Generate Parentheses 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: "((()))", "(()())", "(())()", "()(())", "…

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: [ "((()))", "(()())", "(())()", "()(())", "()()()" ] AC…

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>&…

一天一道LeetCode (一)题目 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: "((()))", "(()())", "(())()", "()(())", "(…

［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: [ "((()))", "(()())", "(())()", "()(())", "()()()"…

题目 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: [ "((()))", "(()())", "(())()", "()(())", "()()()" ]…