算法:3[a]2[b1[c]]ef转为aaabcbcef

const s = '3[a]2[b1[c]]ef'
    function change(str) {
     let stack = []
     let currentStr = '' // 目前累积到的字符串
     let currentNum = 0 // 当前项重复次数
     for(let s of str) {
      if(!isNaN(s)) {
        currentNum = currentNum * 10 + s
      } else if(s === '[') { // 遇到[,先保存以前累积的字符串,和下一次的重复次数
        stack.push(currentStr)
        stack.push(currentNum)
        currentNum = 0
        currentStr = ''
      } else if(s === ']') { // 遇到],弹出以前累积的字符串和,下一次重复次数,拼接
        const prevNum = stack.pop()
        const prevStr = stack.pop()
        currentStr = prevStr + currentStr.repeat(prevNum)
      } else {
        currentStr += s
      }
     }
     return currentStr
    }

    console.log(change(s))