爲什麼我的詞典程序不能打印所有三個字符串？

Question

正如你所看到的，我有三個名字叫做s1，s2和s3。它應該按照字母順序放在新的行上，但「Ashley」不打印。 enter code here

public class Lex { 
static String s1 = "Ashley"; 
static String s2 = "Joe"; 
static String s3 = "John"; 
    public static void main(String[] args) { 
     String topString = ""; 
     String midString = ""; 
     String botString = ""; 

     if (s1.compareTo(s2) > 0 && (s1.compareTo(s3) > 0)) { 
      topString = s1; } 
      else if (s1.compareTo(s2) < 0 && (s1.compareTo(s3) > 0)) { 
      midString = s1; } 
       else{ botString = s1; } 

     if (s2.compareTo(s1) > 0 && (s2.compareTo(s3) > 0)) { 
      topString = s2; } 
      else if (s2.compareTo(s1) < 0 && (s2.compareTo(s3) > 0)) { 
      midString = s2; } 
       else { botString = s2; } 

     if (s3.compareTo(s2) > 0 && (s3.compareTo(s1) > 0)) { 
      topString = s3; } 
      else if (s3.compareTo(s2) < 0 && (s3.compareTo(s1) > 0)) { 
      midString = s3; } 
      else { botString = s3;} 

      System.out.println(topString); 
      System.out.println(midString); 
      System.out.println(botString); 
    } 
} 

Answer 1

你的邏輯複雜，你正在嘗試的東西太多了在同一時間，這將使瞭解&閱讀邏輯就變得更難比較（也是你的代碼縮進不好）。

你可以嘗試下面的代碼與內嵌評論，將採取逐步比較和實現結果。

   if(s1.compareTo(s2) < 0) { //is s1 higher than s2 ? 
       if(s1.compareTo(s3) < 0) { //is s1 higher than s3 as well ? 
        topString = s1;//then topper is s1 
        if(s2.compareTo(s3) < 0) {//is s2 higher than s3 ? 
         midString = s2;//then mid is s2 
         botString = s3;//bot is s3 
        } 
       } else { 
        topString = s3;//s3 is topper compared to s1 and s2 
        midString = s1;//mid is s1 
        botString = s2;//bot is s2 
       } 
      } else {//means s2 is topper 
       if(s2.compareTo(s3) < 0) {//is s2 topper than s3 as well ? 
        topString = s2;//yes, so s2 is topper 
        if(s1.compareTo(s3) < 0) {//is s1 or s3 who is first ? 
         midString = s1;//s1 is first, so at middle 
         botString = s3;//s3 is bot 
        } 
       } else {//s2 is not topper than s3, but topper than s2 
        topString = s3;//top is s3 
        midString = s2;//mid is s2 
        botString = s1;//bot is s1 
       } 
      }