#力扣：20. 有效的括号@FDDLC

20. 有效的括号

一、Java

import java.util.Stack;
 
class Solution {
    public boolean isValid(String s) {
        if (s.length() % 2 == 1) return false;
        Stack stk = new Stack<>();
        for (char c : s.toCharArray()) {
            if (c == ')') {
                if (stk.isEmpty() || stk.pop() != '(') return false;
            } else if (c == ']') {
                if (stk.isEmpty() || stk.pop() != '[') return false;
            } else if (c == '}') {
                if (stk.isEmpty() || stk.pop() != '{') return false;
            } else {
                stk.push(c);
            }
        }
        return stk.isEmpty();
    }
}

import java.util.Stack;
 
class Solution {
    public boolean isValid(String s) {
        if (s.length() % 2 == 1) return false;
        Stack stk = new Stack<>();
        for(char c: s.toCharArray()) {
            if (c == '(') stk.push(')');
            else if(c == '[') stk.push(']');
            else if(c == '{') stk.push('}');
            else if(stk.isEmpty() || stk.pop() != c) return false;
        }
        return stk.isEmpty();
    }
}

import java.util.Stack;
 
class Solution {
    public boolean isValid(String s) {
        if (s.length() % 2 == 1) return false;
        Stack stk = new Stack<>();
        String leftStr = "([{", rightStr = ")]}";
        for (int i = 0; i < s.length(); i++) {
            int idx = rightStr.indexOf(s.charAt(i));
            if (idx != -1) {
                if (stk.isEmpty() || stk.pop() != leftStr.charAt(idx)) return false;
            } else {
                stk.push(s.charAt(i));
            }
        }
        return stk.isEmpty();
    }
}

二、C++

#include 
#include 
 
using namespace std;
 
class Solution {
public:
    bool isValid(string s) {
        if (s.length() % 2 == 1) return false;
        string leftStr = "([{", rightStr = ")]}";
        stack<char> stk;
        for (int i = 0; i < s.length(); i++) {
            auto idx = rightStr.find(s[i]);  // 这里使用auto更好
            if (idx != string::npos) {
                if (stk.empty() || stk.top() != leftStr[idx]) return false;
                stk.pop();
            } else {
                stk.push(s[i]);
            }
        }
        return stk.empty();
    }
};

三、Python

class Solution:
    def isValid(self, s: str) -> bool:
        if len(s) % 2 == 1:
            return False
        stk = []
        m = {')': '(', ']': '[', '}': '{'}
        for c in s:
            if c in ')]}':
                if not stk or stk.pop() != m[c]:  # not stk: len(stk) == 0
                    return False
            else:
                stk.append(c)
        return not stk

四、JavaScript

var isValid = function (s) {
    if (s.length % 2 === 1) return false;
    let stk = [];
    let m = {')': '(', ']': '[', '}': '{'};
    for (let i = 0; i < s.length; i++) {
        if (')]}'.indexOf(s[i]) !== -1) {
            if (stk.length === 0 || stk.pop() !== m[s[i]]) return false;
        } else {
            stk.push(s[i]);
        }
    }
    return stk.length === 0;
};

五、Go

package main
 
import (
	"container/list"
	"strings"
)
 
func isValid(s string) bool {
	if len(s)%2 == 1 {
		return false
	}
	stk := list.New()
	m := map[uint8]uint8{')': '(', ']': '[', '}': '{'}
	for i := 0; i < len(s); i++ {
		if strings.ContainsRune(")]}", rune(s[i])) {
			if stk.Len() == 0 || stk.Remove(stk.Back()) != m[s[i]] {
				return false
			}
		} else {
			stk.PushBack(s[i])
		}
	}
	return stk.Len() == 0
}