Question

我在完成这段代码时遇到了一些困难。基本上，我必须计算一个字母在给定字符串中出现的次数。例如，ABA应输出以下内容

"A appears 1 times"
"B Appears 1 times"
"A Appears 1 times"

但是，我编写的以下代码执行以下操作（它是方法的一部分）

public static char[] counter(char[] original, char[] manipulated) {
    int counter =0;
        for (int i=0; i<manipulated.length; i++) {
            for (int j=0; j<original.length; j++) {
                if (original[j] == manipulated[i]) {
                    counter++;
                } else {

                }
            }
            System.out.println(manipulated[i] + " appears " + counter + " times");
            counter = 0;
        }

    return manipulated;
}

输出是这样的：

"A appears 2 times"
"B appears 1 times"
"A appears 2 times"

哪个没错，但这不是我想要的。那么请你尽快帮我解决这个问题。我知道我想重置一些变量，但我不知道在哪里重置它。提前致谢。

*一些说明：被操纵的变量只是不包含重复的字符串原来是abaa和操纵将是aba :)）

Answer 1

如果你的预期行为只是＆＃34;打印出每个角色出现在一行中的次数＆＃34;那么我不确定你为什么需要manipulated变量。您当前的代码只接受它作为参数，然后将其保持不变。

因此，除非我误解了这个问题，否则它可能是一个更为简单的问题：

void showConsecutiveCharacterCounts(char[] input) {
    int consecutiveCount = 0;
    /* iterate through all chars */
    for (int i = 0; i < input.length; i++) {
        /* increment count - will be 1 first time */
        consecutiveCount++;
        /* if we are at the end of a sequence of chars
           I.e. end of input OR next char does not match */
        if (i == input.length - 1 || input[i] != input[i + 1]) {
            /* print previous sequence */
            System.out.println(input[i] + " appears " + consecutiveCount + " times");
            /* and reset count */
            consecutiveCount = 0;
        }
    }
}

Answer 2

如果您想知道同一个字符连续出现的次数，可以将代码更改为

  public static void main(String[] args) {
    System.out.println(counter("AAACCAAAAAAAB"));
  }

  public static String counter(String s) {
    int counter = 0;
    int j;
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < s.length(); ) {
      j = i;
      char c = s.charAt(i);
      for (; j < s.length(); j++) {
        if (counter == 0) {
          sb.append(c);
        }
        if (c == s.charAt(j)) {
          counter++;
        } else {
          break;
        }
      }
      System.out.println(s.charAt(i) + " appears " + counter + " times");
      counter = 0;
      i = j;
    }
    return sb.toString();
  }

UPDATE 更改了代码，因此您无需提供两个char [] - 现在您输入任何字符串，如＆＃34; AAACCAAAAAAAB＆＃34; （aka＆＃39; original＆＃39;）和char []以前被称为＆＃39;操纵＆＃39;将由该方法返回。

结果： A appears 3 times C appears 2 times A appears 7 times B appears 1 times ACAB

数字信件

2 个答案: