Question

我一直在进行一些编码练习，但我遇到了这个问题。＆＃34;给定一个字符串，返回最大＆＃34;块的长度＆＃34;在字符串中。块是相同的相邻字符序列。＆＃34; 要求输出：

maxBlock("hoopla") → 2
maxBlock("abbCCCddBBBxx") → 3
maxBlock("") → 0

我的代码似乎通过了所有测试，除了最后一个＆＃34;其他测试＆＃34;。有人可以查看我的代码并告诉我哪里可能出错了。

提交代码：

   public int maxBlock(String str) {
      int charBlock = 0;
      int holder = 1;
      if(str.length() == 0){ //If string is empty return 0
         charBlock = 0;
      } else if(str.length() == 1){ //If string contains only a single char return 1
         charBlock = 1;
      } else {
          for(int i=0; i < str.length()-1; i++){   //loop through each char value
              if((str.length() == 2) && (str.charAt(i) != str.charAt(i+1))){ 
                  charBlock =1; //return 1 if the length of the string is 2 and non of the two chars match
              }   
              else if((str.length() == 3) && (str.charAt(i) != str.charAt(i+1))){
                 charBlock = 1; //return 1 if the length of the string is 3 and non of the three chars match
              } 
              else if (str.charAt(i) == str.charAt(i+1)){
                 holder = holder + 1; 
                   if(holder > charBlock){
                     charBlock = holder; //update the value of charBlock if a holder is larger current value
                     }
             } else holder = 1; 
          }
      }
      return charBlock;
    }

Expected    Run     
maxBlock("hoopla") → 2  2   OK      
maxBlock("abbCCCddBBBxx") → 3   3   OK      
maxBlock("") → 0    0   OK      
maxBlock("xyz") → 1 1   OK      
maxBlock("xxyz") → 2    2   OK      
maxBlock("xyzz") → 2    2   OK      
maxBlock("abbbcbbbxbbbx") → 3   3   OK      
maxBlock("XXBBBbbxx") → 3   3   OK      
maxBlock("XXBBBBbbxx") → 4  4   OK      
maxBlock("XXBBBbbxxXXXX") → 4   4   OK      
maxBlock("XX2222BBBbbXX2222") → 4   4   OK      
other tests                             X

Answer 1

这可能更有效，特别是使用辅助方法！不要过分思考这个问题，我知道它很难。我的策略是获取此char的块并转到下一个。

    public int maxBlock(String str) {
       if (str.length() == 0) {
           return 0;
       } 
       int biggest = Integer.MIN_VALUE;
       for (int a = 0; a < str.length(); a++) {
           biggest = Math.max(biggest, (countBlock(str, str.substring(a, a + 1), a));         
       }
       return biggest; //by now we have the biggest
    }

    public int countBlock(String s, String any, int startIndex) {
       int cnt = 1;
       for (int a = startIndex + 1; a < s.length(); a++) { //start right after index
           if (s.substring(a, a + 1).contains(any)){ //keep going
              cnt++;
           }

           else if (!s.substring(a, a + 1).equals(any)) { //stop here
              break;
           }
       }
       return cnt;
    }

Answer 2

Your code is good but you are unnecessarily checking the length of the input string. Here is your code with some improvements.

public int maxBlock(String str) {
    int holder = 1;
    int charBlock = 0;
    for(int i=0; i < str.length()-1; i++){   //loop through each char value
       if (str.charAt(i) == str.charAt(i+1)){
         holder++; 
      }else{
        holder = 1;
      }
      if(holder > charBlock){
         charBlock = holder; //update the value of charBlock if a holder is  larger current value
      }
    }
  return charBlock;
 }

CodingBat maxBlock

2 个答案: