有没有更好的方法来编写 Java 代码来检测字符串中的有效括号? [英] Is there a better way to write a java code for detecting valid parentheses in a String?

问题描述

我试图解决 leetcode 中的一个问题，但我无法通过所有测试用例.我尝试使用堆栈.每当有一个左括号时，我就把它压入堆栈，每当有一个右括号时，我检查并弹出堆栈顶部是否存在正确的左括号.我在最后检查堆栈是否为空

I tried to solve a problem in leetcode and im not able to pass all the test cases. I tried to use stack. Whenever there is an open parenthesis, I push it into the stack and whenever there is a close parentheses, I check and pop if the correct open parenthesis is present on the top of the stack. I check if the stack is empty in the last

问题说明:

给定一个只包含字符 '(', ')', '{', '}', '[' and ']', 确定输入字符串有效.

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

输入字符串在以下情况下有效:

An input string is valid if:

左括号必须由相同类型的括号封闭.
开放括号必须以正确的顺序关闭.
请注意，一个空的字符串也被认为是有效的.

Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.

Stack<Character> stk = new Stack<Character>();
    boolean check = true;
    char temp;
    if(s.length() == 1)
    {
        return false;
    }
    else if(s.length() == 0)
    {
        return true;
    }
    else
    {
        for(int i = 0 ; i < s.length() ; i++)
    {
        temp = s.charAt(i);
        if(temp == '{' || temp == '(' || temp == '[')
        {
            stk.push(temp);
        }
        else if(temp == '}')
        {
            if(stk.peek() == '{')
            {
                stk.pop();
            }
            else
            {
                check = false;
            }
        }
        else if(temp == ')')
        {
            if(stk.peek() == '(')
            {
                stk.pop();
            }
            else
            {
                check = false;
            }
        }
        else if(temp == ']')
        {
            if(stk.peek() == '[')
            {
                stk.pop();
            }
            else
            {
                check = false;
            }
        }
    }
    if(check == false && stk.empty() == false)
    {
        return false;
    }
    else
    {
        return true;
    }
    }

推荐答案

这似乎工作正常:

public boolean isValid(String s) {
    int n = s.length();
    if (n == 0) return true;
    if (n % 2 == 1) return false;

    Stack<Character> st = new Stack<>();
    for (char c : s.toCharArray()) {
        if (c == '{' || c == '[' || c == '(') {
            st.push(c);
        } else if (st.isEmpty()) { // no opening bracket stored
            return false;
        } else {

            char prev = st.pop();
            switch(c) { // matching open and close bracket
                case '}': if (prev != '{') return false; else break;
                case ']': if (prev != '[') return false; else break;
                case ')': if (prev != '(') return false; else break;
            }
        }
    }
    return st.isEmpty(); // stack must be empty in the end - all pairs matched
}

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