针对字谜的数组测试单词-Javascript

Question

到目前为止。我有这个：

function anagrams(word, words) {
  for(let i = 0; i <= words.length; i++){
  const aCharMap = buildCharMap(word);
  const bCharMap = buildCharMap(words[i]);

  if(Object.keys(aCharMap).length !== Object.keys(bCharMap).length) {
    words.pop(words[i])
  }
  for (let char in aCharMap) {
    if (aCharMap[char] !== bCharMap[char]) {
    words.pop(words[i]);
    }
  }
  console.log(word);
  console.log(words);
  }
}

  function buildCharMap(str) {
  const charMap = {};
  for (let char of str.replace(/[^\w]/g, '').toLowerCase()) {
  charMap[char] = charMap[char] + 1 || 1;
  }
  return charMap;
}

如果您阅读了代码，那么眼前的问题就显而易见了，但是这里是

编写一个可以从列表中查找单词的所有字谜的函数。您将获得两个输入一个单词和一个带有单词的数组。您应该返回所有字谜的数组；如果没有，则返回一个空数组。例如：

anagrams('abba', ['aabb', 'abcd', 'bbaa', 'dada']) => ['aabb', 'bbaa']

anagrams('racer', ['crazer', 'carer', 'racar', 'caers', 'racer']) => ['carer', 'racer']

anagrams('laser', ['lazing', 'lazy', 'lacer']) => []

Answer 1

pop是什么？

尝试一下

function anagrams(word, words) {
  let result = [];
  for(let i = 0; i < words.length; i++){
    const aCharMap = buildCharMap(word);
    const bCharMap = buildCharMap(words[i]);

    let test = () => {
  
     if(Object.keys(aCharMap).length !== Object.keys(bCharMap).length) return false;

     for (let char in aCharMap) {
          if (aCharMap[char] !== bCharMap[char]) {
           return false
          }
     }
        
        return true;
  
     }
  
  if(test(aCharMap,bCharMap)) result.push(words[i])
  

  }
  
  function buildCharMap(str) {
    const charMap = {};
    for (let char of str.replace(/[^\w]/g, '').toLowerCase()) {
      charMap[char] = charMap[char] + 1 || 1;
    }
    return charMap;
  }

  return result
}

console.log(anagrams("test",["estt","ttee","tset"]))