我的第一篇简书

这是一道freecodecamp上面的算法题目

用了一堆for循环来解决了这道题，这显然是比较初级的阶段，但是好在自己没有查看其它的资料，独立完成了这个挑战。

题目大概是这样的：
给定两个数组集合，要求返回两个集合的差集。

diffArray([1, 2, 3, 5, 6], [1, 2, 3, 4, 5]) should return an array.

["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"] should return ["pink wool"].

["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"] should return an array with one item.

["andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"] should return ["diorite", "pink wool"].

["andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"] should return an array with two items.

["andesite", "grass", "dirt", "dead shrub"], ["andesite", "grass", "dirt", "dead shrub"] should return [].

["andesite", "grass", "dirt", "dead shrub"], ["andesite", "grass", "dirt", "dead shrub"] should return an empty array.

[1, 2, 3, 5], [1, 2, 3, 4, 5] should return [4].

[1, 2, 3, 5], [1, 2, 3, 4, 5] should return an array with one item.

[1, "calf", 3, "piglet"], [1, "calf", 3, 4] should return ["piglet", 4].

[1, "calf", 3, "piglet"], [1, "calf", 3, 4] should return an array with two items.

[], ["snuffleupagus", "cookie monster", "elmo"] should return ["snuffleupagus", "cookie monster", "elmo"].

assed[], ["snuffleupagus", "cookie monster", "elmo"] should return an array with three items.

[1, "calf", 3, "piglet"], [7, "filly"] should return [1, "calf", 3, "piglet", 7, "filly"].

[1, "calf", 3, "piglet"], [7, "filly"] should return an array with six items.

下面是具体的代码：

function diffArray(arr1, arr2) {
  var newArr1 = [];
  var newArr2 = [];//定义两个数组来执行接下来的运算
  // Same, same; but different.
  arr1.sort();
  arr2.sort();//将输入的两个参数数组排序
  for (let i = 0; i < arr1.length; i++)
  {
    for (let j = 0; j < arr2.length; j++)
    {
      if (arr1[i] === arr2[j])
      {
        newArr1.push(arr1[i]);
      }
    }
  }//求出两个数组的交集newArr1
  for (let i = 0; i < arr1.length; i++)
  {
    var count = 0;//这里使用了一个计数器count实现补集的运算
    for (let j = 0; j < newArr1.length; j++)
    {
      if (arr1[i] === newArr1[j])
      {
        count++;
      }
    }
    if(count === 0)
    {
      newArr2.push(arr1[i]);
    }
  }//求出数组arr1与交集newArr1的补集，并将它所含的元素交给（push到）newArr2
  for (let i = 0; i < arr2.length; i++)
  {
    var count = 0;
    for (let j = 0; j < newArr1.length; j++)
    {
      if (arr2[i] === newArr1[j])
      {
        count++;
      }
    }
    if(count === 0)
    {
      newArr2.push(arr2[i]);
    }
  }//求出数组arr2与交集newArr1的补集，并将它所含的元素交给（push到）newArr2
  return newArr2;//得到最终的差集数组
}

console.log(diffArray([1,2,3,5,6], [1,2,3,4,5]));

可以给这个算法做一个集合运算

已知集合A{x1,x2,x3,...}，B{y1,y2,y3,...}
求 ：C = A && B - A || B
解：令D = A || B，C = （A - D）+（B - D） = A + B - 2D