如何检查字符串中打开和关闭括号的顺序? [英] How to check the sequence of opening and closing brackets in string?

问题描述

需要查找左括号和右括号，如果违反了右括号的顺序，则返回false.

Need to find open and closed bracket, if the sequence of opening and closing brackets is violated, then return false.

但是，如果不还原右数组以与左数组进行比较，则我不在此处做方括号 {[((3 + 1)+2] +} .如果像现在这样反向，那么我无法在此处检查 [1 + 1] +(2 * 2)-{3/3}

But if don't revert right array to compare with left array, i don't make check brackets here {[(3+1)+2]+}. And if reverse like now, then i fail to check here [1+1]+(2*2)-{3/3}

function brackets(expression){
    let leftArr=[];
    let rightArr = [];
    for(let i=0; i<expression.length; i++){
    		if(expression[i] === '(' || expression[i] === '[' || expression[i] === "{"){
        	leftArr.push(expression[i]);
        }
        
        
        if(expression[i] === ')'){
      
        		rightArr.push("(");
        }else if(expression[i] === '}'){
        
        		rightArr.push("{");
        } else if(expression[i] === ']'){
        
        		rightArr.push("[");
        }
   }
		
   rightArr.reverse();
    
    if(leftArr.length<rightArr.length || leftArr.length>rightArr.length){
    return false;
    }
    
    for(let k=0; k<leftArr.length; k++) {
    		if(leftArr[k] != rightArr[k]){
        		return false;
        }
    }

    return true;
}



console.log(brackets('(3+{1-1)}')); // false
console.log(brackets('{[(3+1)+2]+}')); //true
console.log(brackets('[1+1]+(2*2)-{3/3}')); //true
console.log(brackets('(({[(((1)-2)+3)-3]/3}-3)')); //false

推荐答案

在最短的时间内加上注释，可能会使您感到困惑.

In the shortest possible, with comments for lines that are probably confusing for you.

function check(expr){
    const holder = []
    const openBrackets = ['(','{','[']
    const closedBrackets = [')','}',']']
    for (let letter of expr) { // loop trought all letters of expr
        if(openBrackets.includes(letter)){ // if its oppening bracket
            holder.push(letter)
        }else if(closedBrackets.includes(letter)){ // if its closing
            const openPair = openBrackets[closedBrackets.indexOf(letter)] // find its pair
            if(holder[holder.length - 1] === openPair){ // check if that pair is the last element in the array
                holder.splice(-1,1) // if so, remove it
            }else{ // if its not
                holder.push(letter)
                break // exit loop
            }
        }
    }
    return (holder.length === 0) // return true if length is 0, otherwise false
}
check('[[{asd}]]') /// true

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