在Go（Golang）中寻找一个字符串中最长的有效括号子串的程序概述我们的目标是找到一个字符串中最长的有效括号子串的长

概述

我们的目标是找到一个字符串中最长的有效括号子串的长度。

实验结果

Input: ()(()
Output: 2

Input: ()()())
Output: 6

程序

package main

import (
	"fmt"
	"sync"
)

func main() {
	valid := longestValidParentheses("()(()")
	fmt.Println(valid)

	valid = longestValidParentheses("()()())")
	fmt.Println(valid)

	valid = longestValidParentheses("(()")
	fmt.Println(valid)

	valid = longestValidParentheses("(()(()")
	fmt.Println(valid)

	valid = longestValidParentheses("()()")
	fmt.Println(valid)

	valid = longestValidParentheses("(()()")
	fmt.Println(valid)

	valid = longestValidParentheses("((()))(()")
	fmt.Println(valid)

	valid = longestValidParentheses(")(((((()())()()))()(()))(")
	fmt.Println(valid)

	valid = longestValidParentheses("")
	fmt.Println(valid)

}

type customStack struct {
	stack []int
	lock  sync.RWMutex
}

func (c *customStack) Push(name int) {
	c.lock.Lock()
	defer c.lock.Unlock()
	c.stack = append(c.stack, name)
}

func (c *customStack) Pop() (int, error) {
	len := len(c.stack)
	if len > 0 {
		c.lock.Lock()
		defer c.lock.Unlock()
		val := c.stack[len-1]
		c.stack = c.stack[:len-1]
		return val, nil
	}
	return 0, fmt.Errorf("Pop Error: Stack is empty")
}

func (c *customStack) Front() (int, error) {
	len := len(c.stack)
	if len > 0 {
		c.lock.Lock()
		defer c.lock.Unlock()
		return c.stack[len-1], nil
	}
	return 0, fmt.Errorf("Peep Error: Stack is empty")
}

func (c *customStack) Size() int {
	return len(c.stack)
}

func (c *customStack) Empty() bool {
	return len(c.stack) == 0
}

func (c *customStack) Flush() {
	c.stack = []int{}
}

func longestValidParentheses(s string) int {

	customStack := &customStack{
		stack: []int{},
	}

	longestValidParenthesisLength := 0
	currentValidParenthesisLength := 0
	customStack.Push(-1)
	for i, val := range s {

		if val == '(' {
			customStack.Push(i)
		} else if val == ')' {
			customStack.Pop()
			if customStack.Size() == 0 {
				customStack.Push(i)
			} else {
				index, _ := customStack.Front()
				currentValidParenthesisLength = i - index
				if currentValidParenthesisLength > longestValidParenthesisLength {
					longestValidParenthesisLength = currentValidParenthesisLength
				}
			}

		}

	}

	if currentValidParenthesisLength > longestValidParenthesisLength {
		longestValidParenthesisLength = currentValidParenthesisLength
	}

	return longestValidParenthesisLength
}

输出