时间超时，有没有大佬能修改成不超时的（不使用map）

运行超时的js
/**

• lru design

• @param operators int整型二维数组 the ops

• @param k int整型 the k

• @return int整型一维数组

• /
  function LRU( operators , k ) {
  // write code here
  var stack = [];
  var out = [];
  var indexs = [];
  for(var i= 0; i<operators.length;i++){

    var forEachIndex = 0;
    if(operators[i][0]==1){
        if(stack.length>=k){
            var del=0;
            for(var j=0;j<stack.length-1;j++){
  
                if(stack[del][2]<stack[j+1][2]){
                    del=j;
                }else{
                    del=j+1;
                }
            }
            stack.splice(del,1);
        }
        stack.push([operators[i][1],operators[i][2],i]);
        indexs.push(operators[i][1]);
  
    }else if(operators[i][0]==2){
        for(var n = 0;n<stack.length;n++){
            if(stack[n][0]==operators[i][1]){
                //console.log('存在相等项'+i)
                var index = indexs.indexOf(operators[i][1]);
                stack[index][2]=i;
                out.push(stack[n][1]);
                forEachIndex  = forEachIndex + 1;
            }
        }
        if(!forEachIndex){
            out.push(-1);
            //console.log('不存在相等项'+i)
        }
  
    }
    //console.log(i+"次"+stack)
    //console.log(i+"次"+out)

  }
  return out;
  }
  module.exports = {
  LRU : LRU
  };