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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之類的題目,一個一個往資料裡面填,填的時候判斷是否是valid的,每次都要統計括弧數,麻煩。
思路2:改進思路1,也是遞迴實現,將已有的左右括弧數作為參數遞迴下去,寫起來很簡潔。
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++;elsecntR++;}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)); }}
參考:
http://blog.csdn.net/linhuanmars/article/details/19873463
http://blog.csdn.net/fightforyourdream/article/details/14159435