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这可以只用递归来完成吗?

[英]Can this be done with recursion only?

I am stuck on this CodingBat recursion problem:我被困在这个CodingBat 递归问题上:

Given a string, return recursively a "cleaned" string where adjacent chars that are the same have been reduced to a single char.给定一个字符串,递归地返回一个“清理过的”字符串,其中相同的相邻字符已减少为单个字符。 So "yyzzza" yields "yza".所以“yyzzza”产生“yza”。

stringClean("yyzzza") → "yza"
stringClean("abbbcdd") → "abcd"
stringClean("Hello") → "Helo"

I can solve it using loops, but this is not allowed since the problem is supposed so be solved using recursion.我可以使用循环来解决它,但这是不允许的,因为问题应该使用递归来解决。 Is there any way to solve this problem without using a loop and using only recursion?有什么办法可以不使用循环,只使用递归来解决这个问题吗? No global variables, no loops.没有全局变量,没有循环。 I even thought of encoding some information in the parameter but that would be cheating too I think.我什至想过在参数中编码一些信息,但我认为那也是作弊。

My previous program had no while loop, and I could only get half of the answers right.我之前的程序没有while循环,只能答对一半。 Basically, when I called my function with the string parameter, I checked the first 2 characters.基本上,当我用字符串参数调用我的函数时,我检查了前 2 个字符。 If they were the same, I would return the character and call the function again with a string two characters smaller.如果它们相同,我将返回该字符并用一个小两个字符的字符串再次调用该函数。 A string of 3 or 4 of the same consecutive characters would always defeat my algorithm however.然而,由 3 或 4 个相同的连续字符组成的字符串总是会打败我的算法。

public String stringClean(String str) {

    if (str.length() == 0)
        return "";

    if (str.length() > 1) {

    int counter = 1;


      char a = str.charAt(0);
      char b = str.charAt(1);

       if (a == b)
       {
          while (str.length() > 1)
          {
             a = str.charAt(0);
             b = str.charAt(1);

             if (a != b) break;

             counter++;
             str = str.substring(1);


          }

           return a + stringClean( str.substring(1) ) ;
       }

    }

    return str.charAt(0) + stringClean (str.substring(1) );

}

My question is the following, is there any way to solve this problem without using a loop and using only recursion.我的问题如下,有没有办法不使用循环而只使用递归来解决这个问题。 No global variables, no loops.没有全局变量,没有循环。

Answer: Yes.答:是的。 It is very simple.这很简单。 Try below:尝试以下:

 public String stringClean(String str) {
      if (str.length() == 0)
            return "";
      if (str.length() == 1)
            return str;

      if(str.charAt(0) == str.charAt(1)){
         return stringClean(str.substring(1));   
      }else{
        return str.charAt(0)+ stringClean(str.substring(1));
      }    
    }

Your CodingBat results in below:您的 CodingBat 结果如下:

stringClean("yyzzza") → "yza" "yza" OK stringClean("yyzzza") → "yza" "yza" OK
stringClean("abbbcdd") → "abcd" "abcd" OK stringClean("abbbcdd") → "abcd" "abcd" OK
stringClean("Hello") → "Helo" "Helo" OK stringClean("Hello") → "Helo" "Helo" OK
stringClean("XXabcYY") → "XabcY" "XabcY" OK stringClean("XXabcYY") → "XabcY" "XabcY" OK
stringClean("112ab445") → "12ab45" "12ab45" OK stringClean("112ab445") → "12ab45" "12ab45" OK
stringClean("Hello Bookkeeper") → "Helo Bokeper" "Helo Bokeper" OK stringClean("Hello Bookkeeper") → "Helo Bokeper" "Helo Bokeper" OK
other tests OK其他测试正常

My question is the following, is there any way to solve this problem without using a loop and using only recursion.我的问题如下,有没有办法不使用循环而只使用递归来解决这个问题。 No global variables, no loops.没有全局变量,没有循环。

The answer is "yes it is possible".答案是“是的,这是可能的”。

Hints:提示:

  • Like most "tricky" recursive problems this requires an extra parameter.像大多数“棘手”的递归问题一样,这需要一个额外的参数。
  • Think of the problem as filtering the first character of the string at each stage.将问题视为在每个阶段过滤字符串的第一个字符。
  • The first character of the input string is a special case...输入字符串的第一个字符是一个特例......
public String stringClean(String str) {
    if (str == null) {
        return null;
    } else if (str.length() > 1) {
        String k = str.substring(0, 1);
        if (str.charAt(0) == str.charAt(1)) {
            String tmp = stringClean(str.substring(2));
            return k + stringClean(tmp);
        } else {
            return k + stringClean(stringClean(str.substring(1)));
        }
    } else {
        return str;
    }
}

Here is my answer这是我的答案

public String stringClean(String str) {
  if(str.isEmpty()) return "";
  if(str.length()==1)return str;

  if(str.length() > 1 && !str.substring(0,1).equals(str.substring(1,2)))
    return str.substring(0,1) + stringClean(str.substring(1));

  return ""+stringClean(str.substring(1));

}

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