返回重复字母最多的第一个单词

请在下面找到您的代码的可行版本:

function LetterCountI(str) {
    str = str.toLowerCase();
    var arr = str.split(" ");
    var count = 0;
    var word = "-1";
    for (var i = 0; i < arr.length; i++) {
        for (var a = 0; a < arr[i].length; a++) {
            var countNew = 0;
            for (var b = a + 1; b < arr[i].length; b++) {
                if (arr[i][a] === arr[i][b])
                    countNew += 1;
            }
            if (countNew > count) {
                count = countNew;
                word = arr[i];
            }
        }
    }
    return word;
}

您需要将return word;语句移出循环以修复您的版本。

我还把另一种算法放在一起，它依赖于一些内置的javascript方法，如Array.map和Math.max，仅供参考。我做了几次测试，似乎快了几毫秒，但也没快多少。

function LetterCountI(str) {
    var maxCount = 0;
    var word = '-1';
    //split string into words based on spaces and count repeated characters
    str.toLowerCase().split(" ").forEach(function(currentWord){
        var hash = {};
        //split word into characters and increment a hash map for repeated values
        currentWord.split('').forEach(function(letter){
            if (hash.hasOwnProperty(letter)) {
                hash[letter]++;
            } else {
                hash[letter] = 1;
            }           
        });
        //covert the hash map to an array of character counts
        var characterCounts = Object.keys(hash).map(function(key){ return hash[key]; });
        //find the maximum value in the squashed array
        var currentMaxRepeatedCount = Math.max.apply(null, characterCounts);
        //if the current word has a higher repeat count than previous max, replace it
        if (currentMaxRepeatedCount > maxCount) {
            maxCount = currentMaxRepeatedCount;
            word = currentWord;
        }
    });
    return word;
}

这是解决您的问题的Java代码

你的答案不正确。您应该在"for循环"中返回word/Answer/res。查看我的代码。

public static String StringChallenge( String str) { 
  String[] arr = str.split(" ");
  int count = 0; String res = "-1";
  for (int i = 0; i < arr.length ; i++) {
   for (int j = 0; j < arr[i].length() ; j++) {
     int counter = 0;
     for (int k = j + 1; k < arr[i].length() ; k++) {
       if(arr[i].charAt(j) === arr[i].charAt(k) )
          counter ++;
     }
     if (counter > count) {
       count = counter;   res = arr[i];
     }
   }
   return res;
  }
} 

我认为问题是您将return语句放在最外层循环内。

所以你必须把return语句放在内循环中。

正确使用return

     if (countNew > count) {
       count = countNew; 
       word = arr[i];
     }
     return word;
    }
  }
} 

另一个更函数式编程风格的解决方案:

function LetterCountI(str) {
  return ((str = str.split(' ').map(function(word) {
    var letters = word.split('').reduce(function(map, letter) {
          map[letter] = map.hasOwnProperty(letter) ? map[letter] + 1 : 1;
          return map;
        }, {}); // map of letters to number of occurrences in the word
    return {
      word: word,
      count: Object.keys(letters).filter(function(letter) {
        return letters[letter] > 1;
      }).length // number of repeated letters
    };
  }).sort(function(a, b) { // Sort words by number of repeated letters
    return b.count - a.count;
  }).shift()) && str.count && str.word) || -1; // return first word with maximum repeated letters or -1
}
console.log(LetterCountI('Today, is the greatest day ever!')); // => greatest

恰好

http://plnkr.co/edit/BRywasUkQ3KYdhRpBfU2?p =预览

我建议使用正则表达式:/a+/g来查找包含关键字a的字母列表。

我的例子:

var str = aa yyyyy bb cccc cc dd bbb;

首先，找到一个不同的单词列表:

>>> ["a", "y", "b", "c", "d"]

对不同单词列表中的每个单词使用正则表达式:

var word = lstDiffWord[1];
             var
wordcount = str.match(new RegExp(word+'+','g')); 
console.log(wordcount);
>>>>["yyyyy"]