如何计算字符串中的字谜？

Question

我正在尝试解决这个问题：

让函数CountingAnagrams接受str参数并确定字符串中存在多少个字谜。

例子：

输入："aa aa odg dog gdo"?—?输出：2

输入："a c b c run urn urn"?—?输出：1

我试过这个解决方案，但它没有显示正确的答案。我究竟做错了什么？

const CountingAnagrams = (str) => {
  let l = str.length,
    c = 0,
    c1,
    c2;

  if (l % 2 == 0) {
    c1 = str.slice(0, l / 2).split("");
    c2 = str.slice(l / 2).split("");
    
    let l2 = c1.length;

    for (let i = 0; i < l2; i++) {
      let id = c2.indexOf(c1[i]);
      
      if (id !== -1) {
        c2[id] = " ";
      }
      else {
        c += 1;
      }
    }
  }
  else {
    return "-1";
  }
  
  return c;
};

console.log("cars are very cool so are arcs and my os", CountingAnagrams("cars are very cool so are arcs and my os"));
console.log("aa aa odg dog gdo", CountingAnagrams("aa aa odg dog gdo"));
console.log("a c b c run urn urn", CountingAnagrams("a c b c run urn urn"));

Answer 1

const CountingAnagrams = (str) => {
  // Set() helps to remove all the duplicates
  const wordUnique = new Set(str.split(/\s+/)),
    wordArray = [
      ...wordUnique
    ],
    hash = {};
  let count = 0;
  
  
  wordArray.forEach((word) => {
    // Key will be the sorted word e.g. cba will become abc
    let key = word.split('').sort().join('');
    
    // If there is an anagram they will have the same key so whenever the key is avaialable in the hash count will be updated
    if (key in hash) {
      count += 1;
    }
    else {
      // true is assigned just for making the key available in the hash
      hash[key] = true;
    }
  });
  
  return count;
};

console.log("cars are very cool so are arcs and my os", CountingAnagrams("cars are very cool so are arcs and my os"));
console.log("aa aa odg dog gdo", CountingAnagrams("aa aa odg dog gdo"));
console.log("a c b c run urn urn", CountingAnagrams("a c b c run urn urn"));