javascript - 如何检查javascript中的有效大括号，编程问题？

Question

过去几天一直在努力解决代码战中的以下问题:
编写一个接受大括号字符串的函数，并确定大括号的顺序是否有效。它应该返回 true如果字符串有效，并且 false如果它无效。
所有输入字符串都是非空的，并且只包含圆括号、方括号和花括号:()[]{} .
什么被认为是有效的？
如果所有大括号都与正确的大括号匹配，则认为一串大括号是有效的。
例子

"(){}[]"   =>  True
"([{}])"   =>  True
"(}"       =>  False
"[(])"     =>  False
"[({})](]" =>  False

所以我真的坚持使用这个代码，这就是我到目前为止所拥有的:

function validBraces(braces){
    let opening = [ '(', '[', '{']
    let closing = [ ')', ']', '}']
    let count = 0
    const left = []
    const right = []

    // I generate left and right arrays, left w/the opening braces from the input, right w/ the closing
    for (let i = 0; i < braces.length; i++) {
        if (opening.includes(braces[i])) {
            left.push(braces[i])
        } else if (closing.includes(braces[i])) {
            right.push(braces[i])
        }
    }
    if (braces.length % 2 !== 0) {
        return false
    }
    // I know there's no point in doing this but at one point I thought I was finishing the program and thought I would 'optimize it' to exit early, probably this is dumb haha.
    if (left.length !== right.length) {
        return false
    }
    // The juicy (not juicy) part where I check if the braces make sense
    for (let i = 0; i < left.length; i++) {
    // If the list are made up of braces like  ()[]{} add one to counter
        if (opening.indexOf(left[i]) === closing.indexOf(right[i])) {
            count += 1
        } else // If left and right are mirrored add one to the counter
            if (opening.indexOf(left[i]) === closing.indexOf(right.reverse()[i])) {
                count += 1
            }
}
    //If the counter makes sense return true
    if (count === braces.length / 2) {
        return true
    } else { return false}
}


console.log(validBraces( "()" )) //true
console.log(validBraces("([])")) //true
console.log(validBraces( "[(])" )) //false
console.log(validBraces( "[(})" )) //false
console.log(validBraces( "[([[]])]" )) //true

一些评论:我知道我仍然没有检查这个例子 ([])() 但我想以某种方式将它分成两个较小的检查。
谢谢你读到这里。尽管我不希望为我解决问题，但我会以某种方式感谢指导。由于它是一个 6kyu 问题，我可能在某种程度上过于复杂了，如果是这样，关于如何更聪明地处理它的提示将非常感激。
先感谢您! :祈祷::祈祷:

Answer 1

hell 耶!!我很高兴最终自己使用这里给我的一些提示找到了解决方案:

function validBraces(braces){
    let opening = [ '(', '[', '{']
    let closing = [ ')', ']', '}']
    let arr = []
    //console.log(closing.indexOf(braces[")"]) === opening.indexOf(arr[")"]))
    for (let i = 0; i < braces.length; i++) {
        if (opening.includes(braces[i])) {
            arr.push(braces[i])
        } else
        if (closing.indexOf(braces[i]) === opening.indexOf(arr[arr.length - 1])) {
            arr.pop()
        } else return false
    } return arr.length === 0;
}

显然我一开始就想多了哈哈。感谢所有帮助过的人!