[LeetCode in Python] 301 (H) remove invalid parentheses 删除无效的括号

题目

https://leetcode-cn.com/problems/remove-invalid-parentheses/

删除最小数量的无效括号，使得输入的字符串有效，返回所有可能的结果。

说明: 输入可能包含了除?(?和?)?以外的字符。

示例 1:

输入: "()())()"
输出: ["()()()", "(())()"]

示例 2:

输入: "(a)())()"
输出: ["(a)()()", "(a())()"]

示例 3:

输入: ")("
输出: [""]

解题思路

• 先扫描s，确认最少需要删除多少个左括号和右括号，记为left和right
• 然后DFS，对每一个括号都看一下，如果将其干掉，是否可行？
• DFS函数的套路是：
  • 先是退出判断：left和right都为0，且字符串合法
  • 然后从上一次处理的位置开始遍历
  • 如果是括号
    • 如果与前一个字符相同，跳过
    • 因为这会造成重复结果，所以跳过是个优化
  • 如果是左括号，且left>0，则干掉该括号，然后递归DFS，参数中left--
  • 如果是右括号，且right>0，则干掉该括号，然后递归DFS，参数中right--
• 没有重复括号判断也可以工作，只是效率低

代码

class Solution:
    def removeInvalidParentheses(self, s: str) -> List[str]:
        # - find how many left and right parentheses should remove
        left, right = 0, 0
        for c in s:
            if c == ‘(‘:
                left += 1
            elif c == ‘)‘:
                if left == 0:
                    right += 1
                else:
                    left -= 1
            else:
                pass
        
        # - check if valid
        def is_valid(s):
            level = 0
            for c in s:
                if c == ‘(‘:
                    level += 1
                elif c == ‘)‘:
                    if level == 0:
                        return False
                    else:
                        level -= 1
                else:
                    pass

            return level==0

        # - dfs
        def dfs(s, index, left, right, res):
            """
            from index to find ( or ),
            left and right means how many ( and ) to remove
            """

            # - exit check
            if (left == 0) and (right == 0) and is_valid(s):
                res.append(s)
                return

            for i in range(index, len(s)):
                c = s[i]
                if c in [‘(‘, ‘)‘]:
                    # - if continous ( or ), only use first one
                    if (i > 0) and (c == s[i-1]): continue

                    # - try remove ( or )
                    if (c == ‘)‘) and (right > 0):
                        dfs(s[:i]+s[i+1:], i, left, right-1, res)
                    elif (c == ‘(‘) and (left > 0):
                        dfs(s[:i]+s[i+1:], i, left-1, right, res)
                    else:
                        pass

        # - start here
        res = []
        dfs(s, 0, left, right, res)
        
        return list(set(res))

[LeetCode in Python] 301 (H) remove invalid parentheses 删除无效的括号

原文：https://www.cnblogs.com/journeyonmyway/p/12695337.html