如果不使用hasOwnProperty，那么我们有不正确的元素。例如:

[1,2,3].diff([1,2]); //Return ["3", "remove", "diff"] This is the wrong version

我的版本:

Array.prototype.diff = function(array2)
  {
    var a = [],
        diff = [],
        array1 = this || [];

    for (var i = 0; i < array1.length; i++) {
      a[array1[i]] = true;
    }
    for (var i = 0; i < array2.length; i++) {
      if (a[array2[i]]) {
        delete a[array2[i]];
      } else {
        a[array2[i]] = true;
      }
    }

    for (var k in a) {
      if (!a.hasOwnProperty(k)){
        continue;
      }
      diff.push(k);
    }

    return diff;
  }

ES2015的函数方法

计算两个数组之间的差值是Set操作之一。这个术语已经表明应该使用本机Set类型，以便提高查找速度。不管怎样，当你计算两个集合之间的差值时，有三种排列:

[+left difference] [-intersection] [-right difference]
[-left difference] [-intersection] [+right difference]
[+left difference] [-intersection] [+right difference]

下面是反映这些排列的功能性解决方案。

离开的区别:

// small, reusable auxiliary functions const apply = f => x => f(x); const flip = f => y => x => f(x) (y); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); // left difference const differencel = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? false : true ) (xs); }; // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; // run the computation console.log( differencel(xs) (ys) );

正确的区别:

差异是微不足道的。这与翻转的参数不同。为了方便，你可以写一个函数:const differencer = flip(difference)。这是所有!

对称的区别:

现在我们有了左边和右边，实现对称的差异也变得微不足道:

// small, reusable auxiliary functions const apply = f => x => f(x); const flip = f => y => x => f(x) (y); const concat = y => xs => xs.concat(y); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); // left difference const differencel = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? false : true ) (xs); }; // symmetric difference const difference = ys => xs => concat(differencel(xs) (ys)) (flip(differencel) (xs) (ys)); // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; // run the computation console.log( difference(xs) (ys) );

我想这个例子是一个很好的起点，可以让你了解函数式编程的含义:

使用可以以许多不同方式组合在一起的构建块进行编程。

使用额外的内存来做到这一点。这样你可以用更少的时间复杂度来求解，O(n)而不是O(n *n)

function getDiff(arr1,arr2){
let k = {};
let diff = []
arr1.map(i=>{
    if (!k.hasOwnProperty(i)) {
        k[i] = 1
    }
}
)
arr2.map(j=>{
    if (!k.hasOwnProperty(j)) {
        k[j] = 1;
    } else {
        k[j] = 2;
    }
}
)
for (var i in k) {
    if (k[i] === 1)
        diff.push(+i)
}
return diff
}
getDiff([4, 3, 52, 3, 5, 67, 9, 3],[4, 5, 6, 75, 3, 334, 5, 5, 6])

const difference = function (baseArray, arrayToCampare, callback = (a, b) => a!== b) {
  if (!(arrayToCampare instanceof Array)) {
    return baseArray;
  }
  return baseArray.filter(baseEl =>
    arrayToCampare.every(compareEl => callback(baseEl, compareEl)));
}

如果你的数组包含对象，如果你想比较一个属性，就会变得有点困难。

幸运的是，lodash使用_contains和_.pluck使这非常简单:

var list1 = [{id: 1},{id: 2}];
var list1 = [{id: 1},{id: 2}, {id: 3}];

//es6
var results = list2.filter(item => {
  return !_.contains(_.pluck(list1, 'id'), item.id);
});

//es5
var results = list2.filter(function(item){
  return !_.contains(_.pluck(list1, 'id'), item.id);
});

//results contains [{id: 3}]

如果不使用hasOwnProperty，那么我们有不正确的元素。例如:

[1,2,3].diff([1,2]); //Return ["3", "remove", "diff"] This is the wrong version

我的版本:

Array.prototype.diff = function(array2)
  {
    var a = [],
        diff = [],
        array1 = this || [];

    for (var i = 0; i < array1.length; i++) {
      a[array1[i]] = true;
    }
    for (var i = 0; i < array2.length; i++) {
      if (a[array2[i]]) {
        delete a[array2[i]];
      } else {
        a[array2[i]] = true;
      }
    }

    for (var k in a) {
      if (!a.hasOwnProperty(k)){
        continue;
      }
      diff.push(k);
    }

    return diff;
  }