javascript - A front-end interview question, solution

Question

题目： Please give a function to check matching pairs of braces, parenthese and brackets

function isMatchingPair(str) {     
    // your code here 
}
 isMatchingPair('(str[x)xx]')  // return false 
 isMatchingPair('({[str]})')  // return true

Answer 1

function isMatchingPair(str) {     
    var s = [];
    var l = ['(','[','{'];
    var r = [')',']','}'];
    for(var i = 0; i< str.length; i++){
        if(l.includes(str[i]))
            s.push(r[l.indexOf(str[i])]);
        if(r.includes(str[i]))
            if(s.pop()!=str[i]){return false;}
    }
    return s.length?false:true;
}

Answer 2

Stack structure bracket matching.

Maintain a stack structure, traverse the string, and compare it with the current top of the stack. If the top of the stack is left, the traverser is right, and the brackets are of the same type, they are eliminated. Non-parentheses are ignored.

After traversing, if the stack length is 0, it means there is a match, otherwise it does not match.

Answer 3

 function isMatchingPair(str){
      var left = /\(|\{|\[/,right = /\)|\}|\]/,map={'(':')','{':'}','[':']'}
      var stack = []
      var ret = true
      str.split('').forEach(function(e){
        if(left.test(e)){
          stack.push(e)
        }else if(right.test(e)){
          if(e === map[stack[stack.length - 1]]){
            stack.pop()
          }else{
            ret = false
          }
        }
      })
      return ret && !stack.length
    }

Answer 4

// 主要考察的数组的栈操作
function isMatchingPair(str) {    
      let len = str.length; 
      var arr1 = []; // 左括号形成的数组
      let arr2 = []; // 右括号形成的数组
      let obj = {
        '[': ']',
        '(': ')',
        '{': '}'
      };
      const reg1 = /[\[\(\{]+/gi;
      const reg2 = /[\]\)\}]+/gi;
      for (let i = 0; i < len; i++) {
        if (reg1.test(str.charAt(i))) {
          arr1.push(str.charAt(i))
        } else if (reg2.test(str.charAt(i))) {
          arr2.push(str.charAt(i))
        }
      }
      console.log(arr1, arr2);
      if (arr1.length != arr2.length) {
        console.log(false);
        return false;
      }

      for (let i = 0, len = arr1.length; i < len; i++) {
        console.log(i, arr1, arr2);
        if (obj[arr1.shift()] != arr2.pop()) {
          console.log(false);
          return false;
        }
      }
      console.log(true);
      return true;
    }

    isMatchingPair('(str[x)xx]');  // false 
    isMatchingPair('({[str]})');  // true
    isMatchingPair('((((str[[[x))xx])))');  // false 
    isMatchingPair('(({{[[str]]}}))');  // true

Answer 5

Wow, can you ask where the interview questions are? It still feels not easy