Question

在网上看到这个白板挑战，似乎无法弄明白。帮助！

创建一个接受一个单词数组的函数作为它的输入。

你的函数应该返回一个所有单词的数组，这些单词可以使用字母表中的字母输入，只能在标准的美国QWERTY键盘的一行上访问。

例如：

// given
let words = [ 'sup', 'dad', 'tree', 'snake', 'pet'];
keyboardWords(words);

// return
['dad', 'tree', 'pet'];

这就是我已经走了多远。

const topKeys = ['q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p'];
const middleKeys = ['a', 's', 'd','f', 'g', 'h', 'j', 'k', 'l'];
const buttomKeys = ['z', 'x', 'c', 'v', 'b', 'n', 'm'];

let result = [];

let words = ['sup', 'dad', 'tree', 'snake', 'pet'];

for(let i in words) {
  let eachWord = words[i];

  eachWord.split('').forEach(function(c) {
    console.log(c);
  });

}

我已经达到了打印数组中每个单词的程度，但是我不知道用什么方法来查看单个数组中单词中的每个字母是否为topKeys，中键等...

Answer 1

有关详细信息，请参阅Array.prototype.filter()，Set，Spread Syntax，String.prototype.toLowerCase()和Array.prototype.every()。

＆＃13;

// Input.
const input = [
  'ERUT', // top match
  'wdvrmg', // false
  'dsjf', // middle match
  '!^#@&^#', // false
  'CxmvN', // bottom match
  '53454', // false
  '' // false
]

// Match.
const match = (A, B) => [...A].every(x => B.has(x.toLowerCase()))

// Line Words.
const lineWords = words => words.filter(word => word.length && (
  
  // Top match.
  match(word, new Set(['q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p'])) ||
  
  // Middle match.
  match(word, new Set(['a', 's', 'd','f', 'g', 'h', 'j', 'k', 'l'])) ||
  
  // Bottom match.
  match(word, new Set(['z', 'x', 'c', 'v', 'b', 'n', 'm']))
  
))

// Output.
const output = lineWords(input)

// Proof.
console.log(output)

＆＃13;

Answer 2

显然有很多方法可以做到这一点。我最初的反应是制作哈希表。我不确定它比其他任何一个更好，但它应该合理地执行并且易于理解/写在白板上：

＆＃13;

const rows = [
  ['q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p'],
  ['a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l'],
  ['z', 'x', 'c', 'v', 'b', 'n', 'm']
];

// make a quick hash
const hash = rows.reduce((acc, row, i) => (row.forEach(letter => acc[letter] = i + 1), acc), {})
let words = ['sup', 'dad', 'tree', 'snake', 'pet', "4545", "", "PoWer", '0'];

let result = []

words.forEach(word => {
  if (!word) return                     // degenerate cases i.e  '', 0, etc.
  let row = hash[word[0].toLowerCase()]
  if (row == undefined) return          // not letters we know about
  for (let i = 1; i < word.length; i++) {
    if (hash[word[i].toLowerCase()] !== row) return
  }
  result.push(word)
})

console.log(result)

＆＃13;

Answer 3

您应该使用.filter来检查数组的哪些元素通过特定测试。然后使用every查看哪些字词传递：

＆＃13;

const keyboardWords = (() => {
  const topKeys = ['q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p'];
  const middleKeys = ['a', 's', 'd','f', 'g', 'h', 'j', 'k', 'l'];
  const bottomKeys = ['z', 'x', 'c', 'v', 'b', 'n', 'm'];
  return (words) => {
    return words.filter((word) => {
      const chars = word.split('');
      return chars.every(char => topKeys.includes(char))
        || chars.every(char => middleKeys.includes(char))
        || chars.every(char => bottomKeys.includes(char))
    });
  };
})();

const input = [ 'sup', 'dad', 'tree', 'snake', 'pet'];
console.log(keyboardWords(input))

＆＃13;

Answer 4

您可以采用的众多方法之一：

＆＃13;

const topKeys = ['q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p'];
const middleKeys = ['a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l'];
const bottomKeys = ['z', 'x', 'c', 'v', 'b', 'n', 'm'];

const keysets = [topKeys, middleKeys, bottomKeys];

function fn(words) {
	let result = [];

	for (let j = 0; j < words.length; j++) {
		let word = words[j];
		keysets.forEach(function (keyset) {
			if (test(word, keyset)) {
				result.push(word);
			}
		});
	}

	function test(word, keyset) {
		let ok = false;
		for (let i = 0; i < word.length; i++) {
			let char = word.charAt(i);
			ok = keyset.indexOf(char) > -1;
			if (!ok) {
				break;
			}
		}

		return ok;
	}

	return result;
}

let words = ['sup', 'dad', 'tree', 'snake', 'pet'];

let result = fn(words);

console.log(result);

＆＃13;

创建一个接受一系列单词的函数，因为它输入

4 个答案: