求助：编写QWERTY单键盘行可输入单词筛选函数（附代码进度）

Got it, let's work through this problem together. You mentioned you started defining the top keyboard keys, so let's build out the rest of the solution step by step.

Step 1: Define the Keyboard Rows #

First, we need to map out the three rows of a standard QWERTY keyboard. Using Set objects here is ideal because checking if a character exists in a Set is faster (O(1) time) than using an array's includes() method.

const topRow = new Set('qwertyuiop');
const middleRow = new Set('asdfghjkl');
const bottomRow = new Set('zxcvbnm');

Step 2: Create a Helper Function to Check a Single Word #

We'll write a small helper function that takes a word and checks if all its characters belong to the same keyboard row. We'll convert the word to lowercase first to handle case-insensitive inputs (like if a word has uppercase letters).

function isSingleRowWord(word) {
  const lowerWord = word.toLowerCase();
  // Determine which row the first character is in
  let targetRow;
  if (topRow.has(lowerWord[0])) {
    targetRow = topRow;
  } else if (middleRow.has(lowerWord[0])) {
    targetRow = middleRow;
  } else {
    targetRow = bottomRow;
  }
  // Check every character in the word against the target row
  for (const char of lowerWord) {
    if (!targetRow.has(char)) {
      return false;
    }
  }
  return true;
}

Step 3: Build the Main Function #

Now, the main function will take the input array of words, filter out any words that don't pass our helper function's check, and return the valid ones.

function keyboardWords(words) {
  const topRow = new Set('qwertyuiop');
  const middleRow = new Set('asdfghjkl');
  const bottomRow = new Set('zxcvbnm');

  function isSingleRowWord(word) {
    const lowerWord = word.toLowerCase();
    let targetRow;
    if (topRow.has(lowerWord[0])) {
      targetRow = topRow;
    } else if (middleRow.has(lowerWord[0])) {
      targetRow = middleRow;
    } else {
      targetRow = bottomRow;
    }
    for (const char of lowerWord) {
      if (!targetRow.has(char)) {
        return false;
      }
    }
    return true;
  }

  // Filter the input array to keep only valid words
  return words.filter(isSingleRowWord);
}

Test It With Your Example #

Let's run your sample input to verify:

const words = ['sup', 'dad', 'tree', 'snake', 'pet'];
console.log(keyboardWords(words)); // Output: ['dad', 'tree', 'pet']

This matches exactly what you expected!

Optional Optimization #

If you want to make the code a bit more concise, you can combine the row checks into a single array of Sets, then check if all characters in the word are contained in one of them:

function keyboardWords(words) {
  const rows = [
    new Set('qwertyuiop'),
    new Set('asdfghjkl'),
    new Set('zxcvbnm')
  ];
  return words.filter(word => {
    const lowerWord = word.toLowerCase();
    return rows.some(row => {
      return [...lowerWord].every(char => row.has(char));
    });
  });
}

This version does the same thing but uses array methods (some and every) to simplify the logic.

内容的提问来源于stack exchange，提问作者Yaqub