Summary of javascript array deduplication methods_javascript skills

Summary of JavaScript array deduplication methods

Array.prototype.unique1 = function () {
 var n = []; //一个新的临时数组
 for (var i = 0; i < this.length; i++) //遍历当前数组
 {
  //如果当前数组的第i已经保存进了临时数组，那么跳过，
  //否则把当前项push到临时数组里面
  if (n.indexOf(this[i]) == -1) n.push(this[i]);
 }
 return n;
};
Array.prototype.unique2 = function()
{
  var n = {},r=[]; //n为hash表，r为临时数组
  for(var i = 0; i < this.length; i++) //遍历当前数组
  {
    if (!n[this[i]]) //如果hash表中没有当前项
    {
      n[this[i]] = true; //存入hash表
      r.push(this[i]); //把当前数组的当前项push到临时数组里面
    }
  }
  return r;
};
Array.prototype.unique3 = function()
{
  var n = [this[0]]; //结果数组
  for(var i = 1; i < this.length; i++) //从第二项开始遍历
  {
    //如果当前数组的第i项在当前数组中第一次出现的位置不是i，
    //那么表示第i项是重复的，忽略掉。否则存入结果数组
    if (this.indexOf(this[i]) == i) n.push(this[i]);
  }
  return n;
};
Array.prototype.unique4 = function()
{
  this.sort();
  var re=[this[0]];
  for(var i = 1; i < this.length; i++)
  {
    if( this[i] !== re[re.length-1])
    {
      re.push(this[i]);
    }
  }
  return re;
};
var arr = [1,2,2,2,3,3,4,5];
console.log(arr.unique1()); // [1, 2, 3, 4, 5]
console.log(arr.unique2()); // [1, 2, 3, 4, 5]
console.log(arr.unique3()); // [1, 2, 3, 4, 5]
console.log(arr.unique4()); // [1, 2, 3, 4, 5]

The first and third methods both use the indexOf method of the array. The purpose of this method is to find the first occurrence of the stored parameter in the array. Obviously, the js engine will traverse the array until it finds the target when implementing this method. So this function wastes a lot of time. The second method uses a hash table. Store the occurrences in an object in the form of subscripts. Subscripted references are much faster than searching the array using indexOf.

The idea of ​​the fourth method is to sort the array first, and then compare two adjacent values. The JS native sort method is used when sorting. The JS engine should use quick sort internally. The final test result is that the running time of this method is about three times that of the second method on average, but it is much faster than the first and third methods.

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