Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. Have you met this question in a real interview?     Example Given n = 3, a solution set is: "((()))", "(()())", "(())()",…

9.6 Implement an algorithm to print all valid (e.g., properly opened and closed) combinations of n-pairs of parentheses.EXAMPLEInput: 3Output: ((())), (()()), (())(), ()(()), ()()() LeetCode上的原题,请参见我之前的博客Generate Parentheses 生成括号. 解法一: class Solution…

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: "((()))", "(()())", "(())()", "()(())", "()()()" 在LeetCo…

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

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 n 对括号,写一个函数生成所有合适的括号组合.比如,给定 n = 3,一个结果为:[  "((()))",  "(()())",  "(())()",  "()(())",  "()()()"]详见:https://leetcode.com/problems/generate-parentheses/description/ 由于字符串只有左括号和右括号两种字符,而且最终结果必定是左括号3个,右括号3个…

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:  [   "( ( ( ) ) )",   "( ( ) ( ) )",   "( ( ) ) ( )",   "( ) ( ( ) )&q…

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…

生成指定个数的括号,这些括号可以相互包括,但是一对括号的格式不能乱(就是配对的一个括号的左括号要在左边,右括号要在右边) 思维就是从头递归的添加,弄清楚什么时候要添加左括号,什么时候添加右括号 有点像二叉树的建立过程 /* 思路是从第一个符号开始添加,只有两种情况,一种是添加左括号,一种是添加右括号 判断好两种添加的条件后向后添加就行: 1.当左边括号不超过括号数n时可以添加左括号 2.当右括号不超过左括号时可以添加右括号 用递归依次向下添加就行 由于这种数据结构比较像二叉树,代码使用二叉树写的…