在数组的每个索引中，对“,”前面的字符进行计数，并返回最高的

Question

我需要从这个数组中获取最长的姓氏并返回它。

String[] names = {"Turner, Brendan", "Savage, Fred", "Zidy, Boop",
                 "Zobie, Brendan", "Flurb, Fred", "Mopeeeeeyyy, Boopertinson"};

到目前为止我有这个；

public static void getLongestSurname(String[] name){
         int i = 0;
         int x = 0;
         int currentLength = 0; 
         int lastLength = 0; 
         String longestName = null;
    for(int j = 0; j < (name.length); j++){
      
          while (name[j].charAt(i) != ',') {
          i++;
          currentLength++;
          }
      System.out.println(i);
      System.out.println("current is"+currentLength);
      i = 0;
      currentLength = 0;
        if ( currentLength > lastLength ){
           longestName = name[i]; 
          }
    }
    
     System.out.println("longest surname should be; "+ longestName);
  }

但它给出的输出是“Longly surname should be；Turner,Brendan”，这不是列表中最长的名字。

Answer 1

在迭代所有名称时，您需要跟踪LongestName，将当前名称的长度与当前LongestString的长度进行比较，而不是与lastLength进行比较

如果currentName的长度大于longestNumber的长度，将更新longestName

解决方案

public static void getLongestSurname(String[] name){
        
        int currentLength = 0; 
        int lastLength = 0; 
        String longestName = null;
        for(int j = 0; j < (name.length); j++){

           // split string by "," to get surname
            String[] strAr = name[j].split(",");
            // length of current surname 
            currentLength = strAr[0].length();
           // compare length of current surname with longest surName                
            if ( currentLength > lastLength ){
                longestName = name[j]; 
                lastLength = strAr[0].length();
            }
        }

        System.out.println("longest surname should be; "+ longestName);
    }