Question

对于这个挑战，我需要找到重复字母数最多的单词。例如，如果我输入Hello，则输出应为2，因为它包含l或No words的{{1}}个字符，且应为-1 }。我将问题分解为：

1）将句子分成words

数组

2）遍历循环中的每个word

3）遍历循环中的每个charcater

如果一个单词包含的字母数量超过任何其他字母，我就会陷入困境。

public static void main(String[] args) {
    Scanner kbd = new Scanner(System.in);
    System.out.println("Enter any sentence or word combination: ");
    String myString = kbd.nextLine();
    String result = "";
    int count = 0;

    String[] words = myString.split("\\s+");
    for(int i = 0; i < words.length; i++) {
        for(int j = 0; j < words[i].length(); j++) {
            for(int k = 1; k < words[i].length(); k++) {
                char temp = words[i].charAt(k);
                if(temp == words[i].charAt(k-1)) {
                    count++;
                }

            }

        }
    }
}

Answer 1

你几乎做到了，我想你正在研究这样的事情：

static int mostFreqCharCount(final String word) {
    final int chars[] = new int[256];

    int max = 0;
    for (final char c : word.toCharArray()) {
        chars[c]++;
        if (chars[c] > chars[max]) // find most repetitive symbol in word
            max = c;
    }
    return chars[max];
}

public static void main(final String[] args) {
    System.out.println("Enter any sentence or word combination: ");

    final Scanner kbd = new Scanner(System.in);
    final String myString = kbd.nextLine();
    kbd.close();

    int maxC = 0;
    String result = "";

    final String[] words = myString.split("\\s+");
    for (final String word : words) {
        final int c = mostFreqCharCount(word);
        if (c > maxC) {
            maxC = c;
            result = word;
        }
    }

    if (maxC > 1) // any word has at least 1 symbol, so we should return only 2+
        System.out.println(result);
}

主要想法 - 计算每个单词最常用符号的数量，并仅在变量maxC和result

中存储最大值

Answer 2

您想要创建一个length = words.length的数组，并将每个单词的最高值存储在其相对索引中：

int counts[] = new int[words.length];
for(int i = 0; i < words.length; i++) {
    for(int j = 0; j < words[i].length(); j++){
            count = 0
            for(int k = k+1; k < words[i].length(); k++){
                if(words[i].charAt(j) == words[i].charAt(k)){
                    count++;
                }
            if(counts[i] < count)
                  counts[i] = count;
      }

}

然后只扫描数组中的最高值n，并返回单词[n]

Answer 3

如果你只需要一个计数最多的单词，你只需要三个变量，一个用于currentBestWord，一个用于currentLargestCount，另一个用于保持单词中的字符数。

int currentLargestCount=0;
String currentBestWord="";
HashMap<String,Integer> characterCount=new HashMap<String,Integer>();
String[] words=myString.split("\\s+");
for(int i=0;i<words.length;i++){
    String w=words[i];
    characterCount=new HashMap<String,Integer>();
    for(int j=0;j<w.length();j++){
        String character=w.charAt(j).toString();
        if(characterCount.containsKey(character)){
            characterCount.put(character,characterCount.get(character)+1);
        }else{
             characterCount.put(character,1);
        }
    }
    // Now get the largest count of this word
    Iterator ir=characterCount.ValueSet().iterator();
    int thiscount=0;
    while(ir.hasNext()){
         int thischaractercount=ir.next();
         if(thiscount<thischaractercount) thiscount=thischaractercount;
    }
    if(thiscount>currentLargestCount){
        currentLargestCount=thiscount;
        currentBestWord=w;
    }
}

找到具有最多重复字母数的单词

3 个答案: