LeetCode 22. Generate Parentheses 生成括号 Python 回溯解法

题目描述

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
给定n对括号，写一个函数来生成成对的括号的所有组合。

For example, given n = 3, a solution set is:
[ “((()))”, “(()())”, “(())()”, “()(())”, “()()()” ]

解法#1

如果左括号还有剩余，则可以放置左括号，如果右括号的剩余数大于左括号，则可以放置右括号。

class Solution:  
    # @param an integer  
    # @return a list of string  
    def generateParenthesis(self, n):  
        res = []  
        self.generate(n, n, "", res)  
        return res  

    def generate(self, left, right, str, res):  
        if left == 0 and right == 0:  
            res.append(str)  
            return  
        if left > 0:  
            self.generate(left - 1, right, str + '(', res)  
        if right > left:  
            self.generate(left, right - 1, str + ')', res)  

解法#2

回溯中每次遍历字符串s寻找右括号，若右括号在位置i，在此之前的字符串左括号数大于右括号数，且前一个字符为左括号，则与前一字符交换。

class Solution(object):
    result_list = []
    def generateParenthesis(self, n):
        global result_list
        def generate(s):
            global result_list
            new_s = s
            left = right = 0
            for i in range(n * 2):
                if s[i] == '(':
                    left += 1
                else:
                    right += 1
                    if left > right and s[i - 1] == '(':
                        new_s = s[:i - 1] + s[i:i + 1] + s[i - 1:i] + s[i + 1:]
                        if new_s not in result_list:
                            result_list += [new_s]
                            generate(new_s)
        s = '(' * n + ')' * n
        result_list = [s]
        generate(s)
        return result_list