CodingBat 同样结束处理字符串

Question

我正在尝试解决来自 CodingBat 的 this 问题 但不明白为什么它不适用于输入字符串“Hello！”。

这是我的代码，下面是我得到的结果。

public String sameEnds(String string) {
        String result;
        int strLen = string.length();
        boolean isOdd = (strLen % 2 != 0);
        String start = string.substring(0, strLen / 2);
        String end;
        if (isOdd) {
            end = string.substring(strLen / 2 + 1);
        } else {
            end = string.substring(strLen / 2);
        }
        int i = 0;
        while (!start.equals(end) && i <= start.length()) {
            start = start.substring(0, start.length() - i);
            end = end.substring(i, end.length());
            i++;
        }
        if (start.equals(end)) {
            result = start;
        } else {
            result = "";
        }
        return result;
    }

Answer 1

您的问题是您既要增加 i 又要使用 start.length()-i。当i 等于1 时，start 变量会缩短一个字符。但是当i 是2 时，start.length() 已经比原来的少了1，现在你减去2 个字符，所以现在你漏掉了一个。 end 变量也是如此。不要同时使用递增的i 和不断变化的字符串长度。

要修复它，请不要更改原始的 start 和 end 变量。做这样的事情：

    String sTmp = start;
    String eTmp = end;
    while (!sTmp.equals(eTmp) && i <= start.length()) {
        sTmp = start.substring(0, start.length() - i);
        eTmp = end.substring(i, end.length());
        i++;
    }
    if (sTmp.equals(eTmp)) {
        result = sTmp;
    } else {
        result = "";
    }
    return result;

Answer 2

您的代码似乎过于复杂。考虑一下：

public String sameEnds(String string) {
   int e = string.length() - 1;                  /* end of string */
   int b = string.length() / 2;                  /* where to start looking for a match */

   while (--b >= 0) {                            /* ran off the front yet? */

       /*
        * Starting just below the center of the string,
        * look for a character which matches the final character
        */
       for ( ; b >= 0; --b) {
          if (string.charAt(e) == string.charAt(b)) break;
       }

       /*
        * found a match to the final character (a possible starting point)
        * compare characters backwards until no match or all matched
        * (temp vars ee and bb walk backwards from e and b respectively)
        *
        *    "|f|r|o|b|o|z|z|Q|Q|Q|f|r|o|b|o|z|z|"
        *                  ^                   ^
        *                  |                   |
        *         <--bb    b          <--ee    e
        */
       for (int ee = e, bb = b; bb >= 0; --bb, --ee) {
          if (string.charAt(bb) != string.charAt(ee)) break;  /* no match */
          if (bb == 0) return string.substring(0, b+1);       /* victory! */
       }
   }
   return new String("");                              /* nothing matched */
}

Answer 3

这是我的解决方案

public String sameEnds(String string) {
  int mid = string.length() / 2;
  String ls = string.substring(mid);
  String result = "";
  int index = -1;
  int i = 0;

  if (string.length() < 2) {
    return "";
  }

  if (string.length() % 2 == 1) {
    i = 1;
  }

  // All we need to do is loop over the second part of the string,
  // find out the correct substring that equals to the origin string which start the index of '0'.

  for (; i < ls.length(); i++) {
    index = string.indexOf(ls.substring(i));
    if (index == 0) {
      result += ls.substring(i);
      break;
    }
  }

  return result;
}

希望它可以帮助你

Answer 4

public String sameEnds(String string) {
  //First half of the string
  String result = string.substring(0, string.length() / 2); 
  for (int i = 0; i < string.length() / 2; i++)
  {
    //check if first half equals second half
    if (result.equals(string.substring(string.length() - result.length())))
        //return result if true
        return result;
    //else remove last character of first half
    result = result.substring(0, result.length() - 1);
  }
  //return empty string if ends are not equal
  return "";
}

Answer 5

public String sameEnds(String string) {
  final int len = string.length();
  String commonEnd = ""; 

  //Return empty for Anything smaller than size 1 or size 1 
  if (len <= 1) return commonEnd;

  //    S T R I N G
  //    0 1 2 3 4 5
  // left = 2, right = 3
  int left = len / 2;
  int right = len % 2 == 0 ? (len / 2) : (len / 2) + 1;

  while (left > 0) {
    //Start by comparing the string cut down in biggest halves and then keep squeezing
    //left and right from there on
    if (string.substring(0, left).equals(string.substring(right, len))) {
      break;
    }
      left--;
      right++;
  }

  return string.substring(0, left);
}

Answer 6

我的解决方案：

public String sameEnds(String str) {

    int middle = str.length()/2;

    for (int i = middle; i >= 0; i--) {
        String left = str.substring(0,i);

        if (str.endsWith(left)) {
            return left;
        } 

        }

      return "";
}