如何检查一个数组是否是 JavaScript 中另一个数组的子集？

Question

假设我有两个数组，

var PlayerOne = ['B', 'C', 'A', 'D'];
var PlayerTwo = ['D', 'C'];

使用javascript检查arrayTwo是否是arrayOne的子集的最佳方法是什么？

原因：我想理清一个井字游戏的基本逻辑，结果卡在了中间。无论如何，这是我的代码...谢谢大家！

var TicTacToe = {


  PlayerOne: ['D','A', 'B', 'C'],
  PlayerTwo: [],

  WinOptions: {
      WinOne: ['A', 'B', 'C'],
      WinTwo: ['A', 'D', 'G'],
      WinThree: ['G', 'H', 'I'],
      WinFour: ['C', 'F', 'I'],
      WinFive: ['B', 'E', 'H'],
      WinSix: ['D', 'E', 'F'],
      WinSeven: ['A', 'E', 'I'],
      WinEight: ['C', 'E', 'G']
  },

  WinTicTacToe: function(){

    var WinOptions = this.WinOptions;
    var PlayerOne = this.PlayerOne;
    var PlayerTwo = this.PlayerTwo;
    var Win = [];

    for (var key in WinOptions) {
      var EachWinOptions = WinOptions[key];

        for (var i = 0; i < EachWinOptions.length; i++) {
          if (PlayerOne.includes(EachWinOptions[i])) {
            (got stuck here...)
          }

        }
        // if (PlayerOne.length < WinOptions[key]) {
        //   return false;
        // }
        // if (PlayerTwo.length < WinOptions[key]) {
        //   return false;
        // }
        // 
        // if (PlayerOne === WinOptions[key].sort().join()) {
        //   console.log("PlayerOne has Won!");
        // }
        // if (PlayerTwo === WinOptions[key].sort().join()) {
        //   console.log("PlayerTwo has Won!");
        // } (tried this method but it turned out to be the wrong logic.)
    }
  },


};
TicTacToe.WinTicTacToe();

Answer 1

解决方法如下：

使用 ES7（ECMAScript 2016）：

const result = PlayerTwo.every(val => PlayerOne.includes(val));

片段：

const PlayerOne = ['B', 'C', 'A', 'D'];
const PlayerTwo = ['D', 'C'];

const result = PlayerTwo.every(val => PlayerOne.includes(val));

console.log(result);

使用 ES5（ECMAScript 2009）：

var result = PlayerTwo.every(function(val) {

  return PlayerOne.indexOf(val) >= 0;

});

片段：

var PlayerOne = ['B', 'C', 'A', 'D'];
var PlayerTwo = ['D', 'C'];

var result = PlayerTwo.every(function(val) {

  return PlayerOne.indexOf(val) >= 0;

});

console.log(result);

---

这里是在下面的评论中回答问题：

我们如何处理重复项？

解决方案：在上面的解决方案中添加检查数组中足够元素数量的准确条件即可：

const result = PlayerTwo.every(val => PlayerOne.includes(val) 
    && PlayerTwo.filter(el => el === val).length
       <=
       PlayerOne.filter(el => el === val).length
);

第一种情况的片段：

const PlayerOne = ['B', 'C', 'A', 'D'];
const PlayerTwo = ['D', 'C'];

const result = PlayerTwo.every(val => PlayerOne.includes(val) 
    && PlayerTwo.filter(el => el === val).length
       <=
       PlayerOne.filter(el => el === val).length
);

console.log(result);

第二种情况的片段：

const PlayerOne = ['B', 'C', 'A', 'D'];
const PlayerTwo = ['D', 'C', 'C'];

const result = PlayerTwo.every(val => PlayerOne.includes(val) 
    && PlayerTwo.filter(el => el === val).length
       <=
       PlayerOne.filter(el => el === val).length
);

console.log(result);

Answer 2

如果你使用的是 ES6：

!PlayerTwo.some(val => PlayerOne.indexOf(val) === -1);

如果您必须使用 ES5，请为 some 函数和 Mozilla documentation 使用 polyfill，然后使用常规函数语法：

!PlayerTwo.some(function(val) { return PlayerOne.indexOf(val) === -1 });

Answer 3

您可以使用这段简单的代码。

PlayerOne.every(function(val) { return PlayerTwo.indexOf(val) >= 0; })

Answer 4

function isSubsetOf(set, subset) {
    return Array.from(new Set([...set, ...subset])).length === set.length;
}

Answer 5

如果 PlayerTwo 是 PlayerOne 的子集，则集合的长度（PlayerOne + PlayerTwo）必须等于集合的长度（PlayerOne）。

var PlayerOne = ['B', 'C', 'A', 'D'];
var PlayerTwo = ['D', 'C'];

// Length of set(PlayerOne + PlayerTwo) == Length of set(PlayerTwo)

Array.from(new Set(PlayerOne) ).length == Array.from(new Set(PlayerOne.concat(PlayerTwo)) ).length

Answer 6

如果你想比较两个数组并考虑顺序，这里有一个解决方案：

  let arr1 = [ 'A', 'B', 'C', 'D' ];
  let arr2 = [ 'B', 'C' ];
  arr1.join().includes(arr2.join()); //true

  arr2 = [ 'C', 'B' ];
  arr1.join().includes(arr2.join()); //false

Answer 7

这对我来说似乎最清楚：

function isSubsetOf(set, subset) {
    for (let i = 0; i < set.length; i++) {
        if (subset.indexOf(set[i]) == -1) {
            return false;
        }
    }
    return true;
}

它还有一个好处，就是一发现非会员就爆发。

Answer 8

这是一个利用 set 数据类型及其has 函数的解决方案。

let PlayerOne = ['B', 'C', 'A', 'D', ],
    PlayerTwo = ['D', 'C', ],
    [one, two] = [PlayerOne, PlayerTwo, ]
        .map( e => new Set(e) ),
    matches = Array.from(two)
        .filter( e => one.has(e) ),
    isOrisNot = matches.length ? '' : ' not',
    message = `${PlayerTwo} is${isOrisNot} a subset of ${PlayerOne}`;
console.log(message)

Out: D,C is a subset of B,C,A,D