计算最常出现的字符

Question

在用户输入的字符串中，我很难使用计数器。代码定位最常见的字符，但我可以将计数器放在哪里，它计算最常见的字符。Java，请使用当前代码。这是最后一个方法。

import java.util.*;

public class newTest {
    public static void main(String[] args) {
        Scanner scnr = new Scanner(System.in);

        System.out.println("Please enter a one line sentence:");
        String words = scnr.nextLine();

        findAlphabetMode(words);

    }

    public static void findAlphabetMode(String input){
        int[] freq = new int[input.length()];
        char mode = input.charAt(0);
        boolean noMode = true;
        int counter = 0;
        int steadyCount = 0;

        char string[] = input.toCharArray();

    for (int i = 0; i < string.length; i++){
     freq[i] = 1;
        for(int j = i + 1; j < string.length; j++){
          if(string[i] == string[j] && string[i] != ' ' && string[i] != 0){
            freq[i]++;
            //counter++;
          }
          if(counter > 1){
              if(counter > steadyCount){
                  steadyCount = counter + 1;
                  counter = 0;
              }
          }
          if(string[i] == string[j]){
            noMode = false;
            string[j] = 0;
            counter++;
          }
        }
    }
    int max = freq[0];

    for(int i = 0; i < freq.length; i++){
      if(max < freq[i]){
        max = freq[i];
        mode = string[i];
        noMode = false;
      }
    }

        if (noMode) {
            System.out.println("Mode: No Mode");
        }
        else {
            System.out.println("The letter " + mode + " occurs " + steadyCount + " times");
        }
    }
}
`

Answer 1

即使没有流，也不能那么复杂：

public static void findAlphabetMode(String input) {
    var freq = new int['z'-'a'+1];
    for (var ch : input.toLowerCase().toCharArray()) {
        if (ch >= 'a' && ch <= 'z') {
            freq[ch-'a'] += 1;
        }
    }
    var max = 0;  // index to maximum freq
    for (var i = 1; i < freq.length; i++) {
        if (freq[i] > freq[max]) {
            max = i;
        }
    }
    var ch = (char) ('a' + max);
    System.out.printf("found %d times %c%n", freq[max], ch);
}

不难将第二个循环合并到第一个循环中：

public static void findAlphabetMode(String input) {
    var freq = new int['z'-'a'+1];
    var max = 0;  // index to maximum freq
    for (var ch : input.toLowerCase().toCharArray()) {
        if (ch >= 'a' && ch <= 'z') {
            if (++ freq[ch-'a'] > max) {
                max = ch - 'a';
            }
        }
    }
    var ch = (char) ('a' + max);
    System.out.printf("found %d times %c%n", freq[max], ch);
}

注：转换为小写后，仅统计字母a-z

Answer 2

您不需要单独递增计数器，因为您已经在计算数组中每个字符的频率，您可以简单地在数组上循环并找到最大出现字符。以下是代码的简化版本：

import java.util.*;

public class newTest {
    public static void main(String[] args) {
        Scanner scnr = new Scanner(System.in);
        System.out.println("Please enter a one line sentence:");
        String words = scnr.nextLine();
        findAlphabetMode(words);
    }

    public static void findAlphabetMode(String input){
        int[] freq = new int[input.length()];
        char mode = input.charAt(0);
        char string[] = input.toCharArray();
        for (int i = 0; i < string.length; i++){
         freq[i] = 1;
         for(int j = i + 1; j < string.length; j++){
            if(string[i] == string[j]){
              freq[i]++;
            }
         }
      }
      int max = freq[0];

      for(int i = 0; i < freq.length; i++){
        if(max < freq[i]) {
         max = freq[i];
         mode = string[i];
       }
     }
     System.out.println("The letter " + mode + " occurs " + max + " times");
    }
}

Answer 3

您不需要嵌套的for循环。这使得您的算法成为O（N^2），即当您的输入字符串长度加倍时，所用的时间增加了四倍。

您可以将输入字符串中的每个字母放入映射中

public static void findAlphabetMode(String input) {
    input.chars()
        .boxed()
        .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()))
        .entrySet()
        .stream()
        .max(Map.Entry.comparingByValue())
        .ifPresentOrElse(
                e -> System.out.printf("The letter %c occurs %d times%n", e.getKey(), e.getValue()),
                () -> System.out.println("No mode"));
}