为什么这个for循环恢复为0并且永远不会到达这个数组大小的末尾

Question

这个问题是指这里的hackerank anagram问题：

由于某种原因，这个 for 循环永远不会完成。它达到 3（从它迭代的数组的长度为 -2），然后又回到 0，我不知道为什么。

const dictionary = ['hack', 'a' , 'rank' , 'khac','ackh']
const query = ['a','nark','bs','hack','stair']

console.log(stringAnagram(dictionary, query))

function stringAnagram(dictionary, query){
    let sortedDictionary=[];
    let sortedQuery = [];
    let alphabetisedWord;
    let sortedWord;
    let anagramsCount = [];
    // sort them
    for(let i = 0; i<dictionary.length-1; i++){
        sortedWord= dictionary[i];
        if(dictionary[i].length > 1){
         sortedWord= sortedWord.split('');
        console.log('sortedWord: ',sortedWord)
            sortedWord= sortedWord.sort();
            console.log('sortedWord: ',sortedWord)
            sortedWord= sortedWord.join('');
            console.log('sortedWord: ',sortedWord)

            sortedDictionary[i] = sortedWord;
        } else {
            sortedDictionary[i] = sortedWord;
        }
        console.log(i, dictionary.length)
    }
    
    for(let i = 0; i<query.length-1; i++){
        alphabetisedWord = query[i];
        if(query[i].length > 1){

        
        alphabetisedWord = alphabetisedWord.split('');
        console.log('alpha : ', alphabetisedWord)
        alphabetisedWord = alphabetisedWord.sort();
        alphabetisedWord = alphabetisedWord.join('');
        }

        var regex = new RegExp("/" + alphabetisedWord + "/", "g");
        console.log(stringAnagram(dictionary, query))
        anagramsCount[i] = sortedDictionary.toString().match(regex).length
        sortedQuery[i] = alphabetisedWord;
    }
    
    
    return anagramsCount;

}

谁能说出这是什么原因？我已经尝试记录所有索引和单词，但我之前用类似的回答方法做了一个类似的问题 - 只是这一次出现了无限循环，我以前从未见过这种情况。

Answer 1

如果一个简单的问题看起来不可能，那是你做错了……也可以自言自语

const dictionary = ['hack', 'a' , 'rank' , 'khac','ackh']
const query = ['a','nark','bs','hack','stair']

console.log(stringAnagram(dictionary, query))

function stringAnagram(dd,qq){
  var j=(x)=>{return JSON.parse(JSON.stringify(x))} //function to ensure an object isn't passed as pointer JUST IN CASE and to that dude critisizing this part, stringifying an already correctly evaluated obj works(therefore parsing that would be parsing a correctly stringified obj)
  var d=j(dd); var q=j(qq)
  //make words in both query and dictionary now have their words in only 1 format(so that indexOf would work like a charm)
  d=d.map(a=>{return a.split``.sort().join``})
  q=q.map(a=>{return a.split``.sort().join``})
  //now onto indexOf logic(returns the FIRST find of what ur looking for in an array)
  var arr=[]
  q.forEach(a=>{
    var i=0
    while(d.indexOf(a)!=-1){
      var y=d.indexOf(a)
      i++;d.splice(y,1)
    }
    arr.push(i)
  })
  return(arr)
}