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：

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

或 ES5 语法：

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

Answer 2

如果您使用的是ES6：

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

如果必须使用ES5，请对some函数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

如果PlayerTwo是PlayerOne的子集，则set的长度（PlayerOne + PlayerTwo）必须等于set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 5

这对我来说最清楚：

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

一旦找到非成员，它还具有爆发的优势。

Answer 6

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

Answer 7

 Bitmap bitmap = BitmapFactory.decodeResource(getApplicationContext().getResources(), R.drawable.changing);
        TextRecognizer textRecognizer = new TextRecognizer.Builder(getApplicationContext()).build();

        if(!textRecognizer.isOperational()){
            Toast.makeText(getApplicationContext(), "", Toast.LENGTH_SHORT).show();
        }
        else
        {
            Frame frame = new Frame.Builder().setBitmap(bitmap).build();

            SparseArray<TextBlock> item = textRecognizer.detect(frame); //Yeh Frame Detect Kr rha he

            StringBuilder sb = new StringBuilder();
            for(int i = 0 ; i<item.size();i++){
                TextBlock myitem = item.valueAt(i);
                sb.append(myitem.getValue());
                sb.append("\n");
            }

Answer 8

如果您要比较两个数组并同时考虑顺序问题，这是一个解决方案：

  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

检查数组是否是另一个数组的子集

8 个答案: