Generate Parentheses
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:
"((()))", "(()())", "(())()", "()(())", "()()()"
思路:深度优先搜索
本体思路:深搜+动态规划
具体过程:首先判断右边是否到3,到3立即保存,然后深搜,如果没有,立即直到最底层,然后退出,看是否左边大于右边,如果是,加右括号
“((()))” — “(()())”—”(())()”— “()(())” — “()()()”
的确有些难以理解
代码:
class Solution { public: //https://leetcode.com/problems/generate-parentheses/ void parensHelper(int left,int right,int n,vector<string>&result,string temp){ if(right==n){ //右边等于n说明满了,插入temp,返回 result.push_back(temp); return; } if(left<n){ //深度优先 temp+='('; parensHelper(left+1,right,n,result,temp); temp.pop_back(); } //到现在,尽管dfs题目做的很少,但是发现每每深度优先搜索之后,一定有“出栈”,方便下一个入口函数 if(left>right){ temp+=')'; parensHelper(left,right+1,n,result,temp); temp.pop_back(); } } // ( ( ( ) ) ) -- ( ( ( ) ) ) -- vector<string> generateParenthesis(int n) { string temp; vector<string> result; parensHelper(0,0,n,result,temp); return result; } };