[leetcode] 22. Generate Parentheses

题目大意

https://leetcode.com/problems/generate-parentheses/description/

22. 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:

["((()))","(()())","(())()","()(())","()()()"
]

解题思路

The idea here is to only add '(' and ')' that we know will guarantee us a solution (instead of adding 1 too many close). Once we add a '(' we will then discard it and try a ')' which can only close a valid '('. Each of these steps are recursively called.

伪代码(递归法)

if len(s) == 2*n:将s加入到输出列表
if 左括号数目 < n:加入左括号
if 右括号数目 < 左括号数目:加入右括号

Java解法

 public List<String> generateParenthesis(int n) {List<String> list = new ArrayList<String>();backtrack(list, "", 0, 0, n);return list;}public void backtrack(List<String> list, String str, int open, int close, int max){if(str.length() == max*2){list.add(str);return;}if(open < max)backtrack(list, str+"(", open+1, close, max);if(close < open)backtrack(list, str+")", open, close+1, max);}

Python解法

class Solution(object):def generateParenthesis(self, n):""":type n: int:rtype: List[str]"""self.output_list = list()self.n = ndef backtrack(s, left, right):if len(s) == n * 2:self.output_list.append(s)returnif left < self.n:backtrack(s + "(", left + 1, right)if right < left:backtrack(s + ")", left, right + 1)backtrack("", 0, 0)return self.output_list

参考：https://leetcode.com/problems/generate-parentheses/discuss/10100/Easy-to-understand-Java-backtracking-solution

转载于:https://www.cnblogs.com/bymo/p/9644896.html