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[leetcode] 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:
"((()))", "(()())", "(())()", "()(())", "()()()"
https://oj.leetcode.com/problems/generate-parentheses/
思路1:自己模仿permutation之类的题目,一个一个往数据里面填,填的时候判断是否是valid的,每次都要统计括号数,麻烦。
思路2:改进思路1,也是递归实现,将已有的左右括号数作为参数递归下去,写起来很简洁。
思路1:
public class Solution { public ArrayList<String> generateParenthesis(int n) { ArrayList<String> result = new ArrayList<String>(); if (n <= 0) return result; n = n * 2; char str[] = new char[n]; generate(0, n, str, result); return result; } private void generate(int cur, int n, char[] s, ArrayList<String> result) { if (cur == n) { result.add(new String(s)); } else { int j; int cntL = 0, cntR = 0; for (j = 0; j < cur; j++) { if (s[j] == ‘(‘) cntL++; else cntR++; } if (cntL > cntR) { s[cur] = ‘)‘; generate(cur + 1, n, s, result); if (cntL < n / 2) { s[cur] = ‘(‘; generate(cur + 1, n, s, result); } } else if (cntL == cntR) { s[cur] = ‘(‘; generate(cur + 1, n, s, result); } else { return; } } } public static void main(String[] args) { System.out.println(new Solution().generateParenthesis(3)); }}
思路2:
import java.util.ArrayList;public class Solution { public ArrayList<String> generateParenthesis(int n) { ArrayList<String> res = new ArrayList<String>(); if (n <= 0) return res; StringBuilder sb = new StringBuilder(); generate(n, n, sb, res); return res; } private void generate(int l, int r, StringBuilder sb, ArrayList<String> res) { if (r < l) return; if (l == 0 && r == 0) { res.add(sb.toString()); } if (l > 0) { sb.append("("); generate(l - 1, r, sb, res); sb.deleteCharAt(sb.length() - 1); } if (r > 0) { sb.append(")"); generate(l, r - 1, sb, res); sb.deleteCharAt(sb.length() - 1); } } public static void main(String[] args) { System.out.println(new Solution().generateParenthesis(3)); }}
参考:
http://blog.csdn.net/linhuanmars/article/details/19873463
http://blog.csdn.net/fightforyourdream/article/details/14159435
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