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比较多个字符串并排序答案

compareto java

0

Ash Dean · 技术社区 · 11 年前

我试图比较3个不同的单词,然后对它们进行排序,但我没有得到正确的答案。一些方向将是一个很大的帮助! 对于Java来说,这是一个非常新的概念,并且已经在这个问题上停留了太久。

到目前为止,我使用的是:

`// word1, word2, and word3 
   /* If word1 is greater than word2, 
    * and word2 IS greater than word3 and word3 is NOT greater then word1: display word1,     word2, word3 
    */
   result1  =   word1.compareTo(word2);
   result1a =   word1.compareTo(word3);
   result1b =   word2.compareTo(word3);

//display and compare information
       if (result1 > result1a + result1b)  
       JOptionPane.showMessageDialog(null, word1 + word2 + word3);

       else if (result1 < result1a + result1b)
       JOptionPane.showMessageDialog(null, word2 + word3 + word1);

`

2 回复 | 直到 11 年前

1

R2B2 11 年前

String message;
if(word2.compareTo(word1) < 0) { //word2 < word1
    if(word3.compareTo(word1) < 0) { //word3 < word1
        if(word3.compareTo(word2) < 0) { //word3 < word2
            message = word3 + word2 + word1;
        } else { //word2 <= word3
            message = word2 + word3 + word1;
        }
    } else { //word1 <= word3
        message = word2 + word1 + word3;
    }
} else { //word1 <= word2
    if(word3.compareTo(word2) < 0) { //word3 < word2
        if(word3.compareTo(word1) < 0) { //word3 < word1
            message = word3 + word1 + word2;
        } else { //word1 <= word3
            message = word1 + word3 + word2;
        }
    } else { //word2 <= word3
        message = word1 + word2 + word3;
    }
}

这是使用词典/字母顺序。如果你想用单词的长度来比较,你必须替换每个单词 s1.compareTo(s2) 通过 s1.length() - s2.length() (如果s1大于s2,则应为正整数)。

2

1

Sagar D 11 年前

ArrayList<String> list = new ArrayList<String>(); //create an arraylist

list.add("ccc")  //add the strings
list.add("aaa");
list.add("bbb");

Collections.sort(list); //this will sort the list in lexographical precedence

//now get the words and process as you want