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Tic tac toe游戏的基本逻辑,结果卡在中间了。这是我的代码。非常感谢!
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();
发布于 2016-08-07 15:44:39
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 });
发布于 2017-05-30 17:08:20
您可以使用这段简单的代码。
PlayerOne.every(function(val) { return PlayerTwo.indexOf(val) >= 0; })
发布于 2019-04-12 11:51:15
function isSubsetOf(set, subset) {
return Array.from(new Set([...set, ...subset])).length === set.length;
}
https://stackoverflow.com/questions/38811421
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