[LeetCode]Generate Parentheses, 解题报告 - c++编程基础

前言

发现做递归的题目的时候还是会遇到一些问题，有时候就怕转不过弯来

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

其实思路就是先放左括号，再放右括号，直到左右括号的数量均==n即可。注意：放括号的过程中，永远不能出现当前右括号的数量大于左括号数量的情况

AC代码

public class Solution {
    public static ArrayList
  
    generateParenthesis(int n) {
        ArrayList
   
     list = new ArrayList
    
     (); StringBuilder str = new StringBuilder(); if (n == 0) { return list; } recursive(0, 0, n, str, list); return list; } public static void recursive(int left, int right, int n, StringBuilder str, ArrayList
     
       list) { if (left < right) { return; } if (left == n && right == n) { String tmp = str.toString(); list.add(tmp); return; } if (left < n) { StringBuilder newstr = new StringBuilder(str.toString()); newstr.append('('); recursive(left + 1, right, n, newstr, list); } if (right < n) { StringBuilder newstr = new StringBuilder(str.toString()); newstr.append(')'); recursive(left, right + 1, n, newstr, list); } } }