标签：reading lin ddp com function problem 链接 air 题目
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:
[ "((()))", "(()())", "(())()", "()(())", "()()()" ]
链接：https://leetcode.com/problems/generate-parentheses/#/description
4/12/2017
3ms, 60%
我认为在我刷题之前这道题我肯定能做出来，现在反而一开始一点思路都没有了。还好这道题可以用backtracking来解决，所以掌握了方法是多么地重要。
注意的问题：
1. addParenthesis当中一定要把case的判断条件写详细。比如从第15行开始，最开始我的判断条件是if (left == n) {...} else {...}，但是结果会溢出。为什么？因为{...}也许会有错误，导致根本不能确保left >= right，这只是我的猜想，应该不是最根本原因
1 public class Solution { 2 public List<String> generateParenthesis(int n) { 3 List<String> ret = new ArrayList<>(); 4 if (n <= 0) return ret; 5 StringBuilder sb = new StringBuilder(); 6 addParenthesis(0, 0, n, ret, sb); 7 return ret; 8 } 9 10 private void addParenthesis(int left, int right, int n, List<String> ret, StringBuilder sb) { 11 if (left == right && left == n) { 12 ret.add(sb.toString()); 13 return; 14 } 15 if (left == right) { 16 sb.append(‘(‘); 17 addParenthesis(left + 1, right, n, ret, sb); 18 sb.deleteCharAt(sb.length() - 1); 19 } else if (left > right) { 20 if (left < n) { 21 sb.append(‘(‘); 22 addParenthesis(left + 1, right, n, ret, sb); 23 sb.deleteCharAt(sb.length() - 1); 24 } 25 sb.append(‘)‘); 26 addParenthesis(left, right + 1, n, ret, sb); 27 sb.deleteCharAt(sb.length() - 1); 28 } 29 } 30 }
前面提到的错误的溢出解法，根本不能保证right比left小，也不能保证right比n小
1 public class Solution { 2 public List<String> generateParenthesis(int n) { 3 List<String> ret = new ArrayList<>(); 4 if (n <= 0) return ret; 5 StringBuilder sb = new StringBuilder(); 6 addParenthesis(0, 0, n, ret, sb); 7 return ret; 8 } 9 10 private void addParenthesis(int left, int right, int n, List<String> ret, StringBuilder sb) { 11 if (left == right && left == n) { 12 ret.add(sb.toString()); 13 return; 14 } 15 if (left == n) { 16 sb.append(‘)‘); 17 addParenthesis(left, right + 1, n, ret, sb); 18 sb.deleteCharAt(sb.length() - 1); 19 } else { 20 sb.append(‘)‘); 21 addParenthesis(left, right + 1, n, ret, sb); 22 sb.deleteCharAt(sb.length() - 1); 23 sb.append(‘(‘); 24 addParenthesis(left + 1, right, n, ret, sb); 25 sb.deleteCharAt(sb.length() - 1); 26 } 27 } 28 }
别人的做法：
思路一致，更好看的写法
https://discuss.leetcode.com/topic/8724/easy-to-understand-java-backtracking-solution
1 public List<String> generateParenthesis(int n) { 2 List<String> list = new ArrayList<String>(); 3 backtrack(list, "", 0, 0, n); 4 return list; 5 } 6 7 public void backtrack(List<String> list, String str, int open, int close, int max){ 8 9 if(str.length() == max*2){ 10 list.add(str); 11 return; 12 } 13 14 if(open < max) 15 backtrack(list, str+"(", open+1, close, max); 16 if(close < open) 17 backtrack(list, str+")", open, close+1, max); 18 }
各种Python写法，更好的是，他提供了升级Python的一些建议：
https://discuss.leetcode.com/topic/17510/4-7-lines-python
I learned a lot by reading tutorials/docs and by looking at solutions of others at checkio.org (a Python programming site) and by spending a month on Python topics at Stackoverflow. Also, practice practice practice :-)
还提到了，我昨天刚研究到
http://docs.python-guide.org/en/latest/writing/gotchas/
其他人还用DP的做法，时间复杂度也许并不好:
https://discuss.leetcode.com/topic/3474/an-iterative-method
更多讨论：
https://discuss.leetcode.com/category/30/generate-parentheses
标签：reading lin ddp com function problem 链接 air 题目
原文地址：http://www.cnblogs.com/panini/p/6701997.html
