检查数组是否是另一个数组的子集

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？

原因是:我试图理清游戏Tic tac toe的基本逻辑,并陷入中间.无论如何,这是我的代码...谢谢堆!

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

正确的解决方案是:

在ES6语法中:

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

或者在ES5语法中:

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

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

console.log(result);

Answer 2

如果您使用的是ES6:

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

如果必须使用ES5,请使用polyfill作为Mozilla文档的some函数,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set（PlayerOne + PlayerTwo）的长度必须等于set（PlayerOne）的长度。

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