有效括号相关

相关题目
20. 有效的括号
921. 使括号有效的最少添加
1541. 平衡括号字符串的最少插入次数
32. 最长有效括号

# 20. 有效的括号
class Solution:
    def isValid(self, s: str) -> bool:
        stack = []
        for pare in s:
            if pare in '([{':
                stack.append(pare)

            if not stack or (pare == ')' and stack[-1] != '('):
                return False

            elif not stack or (pare == ']' and stack[-1] != '['):
                return False

            elif not stack or (pare == '}' and stack[-1] != '{'):
                return False

            if pare in ')]}':
                stack.pop()
                
        return not stack

# 921. 使括号有效的最少添加
class Solution:
    def minAddToMakeValid(self, s: str) -> int:
        # 对左括号的需求
        res = 0
        # 对右括号的需求
        need = 0

        for c in s:
            if c == '(':
                # 右括号需求
                need += 1
            else:
                need -= 1
                if need == -1:
                    need = 0
                    # 需要插入一个左括号
                    res += 1

        return res + need

# 1541. 平衡括号字符串的最少插入次数
class Solution:
    def minInsertions(self, s: str) -> int:
        # 对左括号的需求
        res = 0
        # 对右括号的需求
        need = 0

        for c in s:
            if c == '(':
                # 右括号需求
                need += 2
                if need % 2 == 1:
                    # 插入一个右括号
                    res += 1
                    # 右括号需求减1
                    need -= 1
            else:
                need -= 1
                if need == -1:
                    # 需要插入一个左括号
                    res += 1
                    # 同时右括号的需求变为1
                    need = 1

        return res + need
        

# 32. 最长有效括号
# 动态规划+栈
class Solution:
    def longestValidParentheses(self, s: str) -> int:
        # dp[i] 表示已s[i-1]结尾的最长合法括号子串的长度
        dp = [0] * (len(s) + 1)
        stk = []

        for idx, c in enumerate(s):
            if c == '(':
                # 遇到左括号，记录索引
                stk.append(idx)
                # 左括号结尾，不可能是有效合法括号
                dp[idx+1] = 0

            else:
                if stk:
                    # 配对的左括号对应索引
                    leftIndex = stk.pop()
                    dp[idx+1] = (idx - leftIndex + 1) + dp[leftIndex]

                else:
                    dp[idx+1] = 0

        return max(dp)