递归匹配括号

Question

我写了下面的递归代码来匹配括号。在某些情况下，我得到“True”，但如果我在字符串中的某个位置添加新的括号（这是不正确的），我仍然得到“True”。我调试了一下，不明白如何修复，请问如何修复？

def is_balanced(parenthesis):
    if len(parenthesis) %2 == 1:
        return False
        left_value = parenthesis[:1:]
        right_value = parenthesis[-1::]
        return is_balanced(parenthesis[1:-1:]) if left_value != right_value else 
        True

    print(is_balanced('(()()[]()())')) => #True
    print(is_balanced('(()()[[()())')) => #still True

Answer 1

这是一个相当简洁的基于正则表达式的实现：

import re

def is_balanced(par):
    pattern = re.compile('\(\)|{}|\[\]')  # matches '()', '[]', or '{}'
    return not par or bool(pattern.search(par)) and is_balanced(pattern.sub('', par))

Answer 2

一种递归方法如下：

def is_balanced(parenthesis):
    l = len(parenthesis)
    if l == 0:
        return True
    else:
        if '()' in parenthesis or '[]' in parenthesis:
            return is_balanced(parenthesis.replace('()', '').replace('[]', ''))
        else:
            return False

print(is_balanced('(()()[]()())'))  # True
print(is_balanced('(()()[[()())'))  # False

---

这里的想法是递归地用空字符串替换闭括号和闭括号，然后看看最终是否得到空字符串。

更简单但非递归的方法是：

def is_balanced(parenthesis):
    brackets = ['()', '{}', '[]']
    while any(x in parenthesis for x in brackets):
        for br in brackets:
            parenthesis = parenthesis.replace(br, '')
    return not parenthesis

Answer 3

var inputString = "[[1+1]+(2*2)-{3/3}<}]";

console.log("Input String:", inputString);

//remove non bracket characters
var regexNonBracketPattern = /[^<>{}[\]()]/g;
var bracketOnlyString = inputString.replace(regexNonBracketPattern, "");

//find if it has any matching bracket
console.log("Bracket Only String:", bracketOnlyString);

var regexMatchingBracketsPattern = /(\(\)|\[\]|\<\>|\{\})+/;
let hasAnyMatchingBracket =
regexMatchingBracketsPattern.test(bracketOnlyString);

console.log("Has Any Matching Bracket:", hasAnyMatchingBracket);

var counter = 1;
while (regexMatchingBracketsPattern.test(bracketOnlyString)) {
    bracketOnlyString = bracketOnlyString.replace(regexMatchingBracketsPattern,"");

    if (bracketOnlyString == "") {
        break;
    } else {
        console.log("Counter Value:", counter);
        console.log("Iteration No (" + counter + "):", bracketOnlyString);
    }
    counter++;
}

if (bracketOnlyString.length != 0) {
    console.log("Provided string has some non matching brackets");
} else {
    console.log("Provided string has all matching brackets");
}