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:我试图理清井字游戏的基本逻辑,结果陷入了困境。无论如何,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();
最佳答案
解决方案如下:
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);
这是回答下面评论中的问题:
How do we handle duplicates?
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
);
第一个案例的片段:
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);
关于javascript - JavaScript 中如何检查一个数组是否是另一个数组的子集?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38811421/