LeetCode 22 - 括号生成

题目描述

22. 括号生成

解法一：dfs（Python）

解法一和解法二都可以详细参考 回溯算法（深度优先遍历）+ 广度优先遍历 + 动态规划

class Solution:
    def generateParenthesis(self, n: int) -> List[str]:
        res  = []
        cur_str = ''
        
        def dfs(cur_str, left, right):
            if left == 0 and right == 0:
                res.append(cur_str)
                return 
            if right < left:
                return 
            if left > 0:
                dfs(cur_str + '(', left - 1, right)
            if right > 0:
                dfs(cur_str + ')', left, right - 1)

        dfs(cur_str, n, n)
        return res

解法二：动态规划（Python）

简单的来说就是一句话：

$dp[i] = "(" + dp[可能的括号对数] + ")" + dp[剩下的括号对数]$

class Solution:
    def generateParenthesis(self, n: int) -> List[str]:
        if n == 0: return []
        dp = [0  for _ in range(n + 1)]
        dp[0] = [""]
        for i in range(1, n+1):
            cur = []
            for j in range(i):
                left = dp[j]
                right = dp[i - j - 1]
                for s1 in left:
                    for s2 in right:
                        cur.append("(" + s1 + ")" + s2)
            dp[i] = cur
        return dp[n]