Question

我正在编写来自codingbat.com的递归问题，该问题表明

Given a string, compute recursively a new string where all the lowercase 'x' chars have been moved to the end of the string.

它传递了所有示例，除了“其他测试”的示例。由于我无法看到“其他测试”指的是什么，我被卡住了。任何帮助，将不胜感激。这是我的代码

public String endX(String str) {
  return endX2(str, 0, str.length());
}

public String endX2(String str, int n, int len){
  if(len == 0) return str;
  if(n == len-1) return str;
  if(str.substring(n,n+1).equals("x")){
    return str.substring(0,n) + (endX2(str.substring(n+1) + "x", n, len-1));
  }
  else return endX2(str, n+1, len);
}

Answer 1

我不确定为什么你有一个额外的方法，因为它是不必要的。您还需要额外检查 - if(n == len-1) return str; - 这是不需要的。

您遇到的问题是您使用未经检查的索引，并且当使用的字符串不以＆＃39; x＆＃39;或许多＆＃39; x。如果我针对String xs = "xfooxbarxx";使用您的代码，请为我java.lang.StringIndexOutOfBoundsException。我还没有广泛地调试代码，但是这应该能够理解为什么它在＆＃34;其他＆＃34;上失败了。试验。看看我的版本，自己调查问题所在，以及如何让你的代码更简洁。

public String endX(String str) {
  if(str.length() == 0)
    return "";
  return str.startsWith("x") ? 
    endX(str.substring(1)) + "x" :
    String.valueOf(str.charAt(0)) + endX(str.substring(1));
}

Answer 2

PS：这是比它需要更长的方式。

/*
Alright here is what we need to do...
Step 1: Get all the 'x' chars into a String.
Step 2: Get all NON 'x' chars into a String.
Step 3 (goal): Concencate (combine) the NON 'x' String first then the 'x' String in that order.
Solution Notes: Instead of using an index variable to go through a String, we could 'substring' off the first char in the String each time, cutting until we are down to the base case, for the sake of showing recursive logic I used an index varible. However on Arrays or any other collection, you need an index varible to access that element ant that spot or (index).
*/
public String endX(String str) {
  //Ternary operator used here...could be written as a full if then else statement.
  //Ternary operator logic: if the length is 0 return "", else return getNonX(str, "", 0) + getX(str, "", 0);
  return (str.length() == 0) ? "" : getNonX(str, "", 0) + getX(str, "", 0);

  //NOTICE in the parts [getNonX(str, "", 0)] and [getX(str, "", 0)] 
  //there is an empty String in the middle, that is there to hold or gather the
  //chars, 'x' or not. We fill those empty Strings up in each recursive helper
}
public String getX(String str, String x, int index) {
  //We are at the end, and if the last char IS an 'x'...

  if(index == str.length() - 1 && str.charAt(index) == 'x'){
  return x + 'x'; //...put that last 'x' on the 'x' String.
  }

  //We are at the end and if the last char IS NOT an 'x'...

  else if(index == str.length() - 1 && str.charAt(index) != 'x'){
  return x; //...just return what we got.
  }    
  //When we see an 'x' but we aren't at the end...

  if(index < str.length() - 1 && str.charAt(index) == 'x'){
  x += 'x'; //...append 'x' to the 'x' String.
  }

  return getX(str, x, index + 1); //recur, moving the index up
}
public String getNonX(String str, String nonX, int index) {
  //We are at the end, and if the last char IS NOT an 'x'...

  if(index == str.length() - 1 && str.charAt(index) != 'x'){
  return (nonX + str.charAt(index)); //...append that char to the 'nonX' String
  }

  //We are at the end and if the last char IS an 'x'...

  else if(index == str.length() - 1 && str.charAt(index) == 'x'){
  return nonX; //...just return what we got.
  }

  //When we see a non 'x' char and we aren't at the end...

  if(index < str.length() - 1 && str.charAt(index) != 'x'){
  nonX += str.charAt(index); //...append that char to the 'nonX' String
  }

  return getNonX(str, nonX, index + 1); //recur, move the index up
}

Answer 3

您可以尝试此代码...

public String endX(String str) {
  if(str.length() <=1) return str;
  if(str.charAt(0) == 'x') return endX(str.substring(1)) + 'x';
  return str.charAt(0) + endX(str.substring(1));
}

Codingbat.com endX没有解决＆＃34;其他测试＆＃34;

3 个答案: