为链接列表中的字符编写compareTo（）方法

我在为构建字符串的链接列表对象编写compareTo()和charAt()方法时遇到问题。这个名为LString的类包含一个构造函数和其他一些方法。它运行与测试其能力，作为一个链表字符串生成另一个文件，我收到此错误信息：为链接列表中的字符编写compareTo（）方法

Running constructor, length, toString tests (10 tests) 
Starting tests: .......... 
Time: 0.000 
OK! (10 tests passed.) 

Running compareTo and equals tests (18 tests) 
Starting tests: EEEEEEEEEEEE...... 
Time: 0.015 

There were 12 failures: 
1) t21aTestCompareTo[0](LStringTest$LStringCompareToTest) 
java.lang.AssertionError: compareTo of "abc" and "abd" wrong expected:<-1> but was:<0> 
     at org.junit.Assert.fail(Assert.java:88) 
     at org.junit.Assert.failNotEquals(Assert.java:743) 
     at org.junit.Assert.assertEquals(Assert.java:118) 
     at org.junit.Assert.assertEquals(Assert.java:555) 
     at LStringTest$LStringCompareToTest.t21aTestCompareTo(LStringTest.java:259) 
     ... 9 more 
2) t22aTestEquals[0](LStringTest$LStringCompareToTest) 
java.lang.AssertionError: equals of "abc" and "abd" wrong expected:<false> but was:<true> 
     at org.junit.Assert.fail(Assert.java:88) 
     at org.junit.Assert.failNotEquals(Assert.java:743) 
     at org.junit.Assert.assertEquals(Assert.java:118) 
     at LStringTest$LStringCompareToTest.t22aTestEquals(LStringTest.java:269) 
     ... 9 more 
3) t21aTestCompareTo[1](LStringTest$LStringCompareToTest) 
java.lang.AssertionError: compareTo of "a" and "" wrong expected:<1> but was:<0> 
     at org.junit.Assert.fail(Assert.java:88) 
     at org.junit.Assert.failNotEquals(Assert.java:743) 
     at org.junit.Assert.assertEquals(Assert.java:118) 
     at org.junit.Assert.assertEquals(Assert.java:555) 
     at LStringTest$LStringCompareToTest.t21aTestCompareTo(LStringTest.java:261) 
     ... 9 more 
4) t22aTestEquals[1](LStringTest$LStringCompareToTest) 
java.lang.AssertionError: equals of "" and "a" wrong expected:<false> but was:<true> 
     at org.junit.Assert.fail(Assert.java:88) 
     at org.junit.Assert.failNotEquals(Assert.java:743) 
     at org.junit.Assert.assertEquals(Assert.java:118) 
     at LStringTest$LStringCompareToTest.t22aTestEquals(LStringTest.java:269) 
     ... 9 more 
5) t21aTestCompareTo[2](LStringTest$LStringCompareToTest) 
java.lang.AssertionError: compareTo of "a" and "ab" wrong expected:<-1> but was:<0> 
     at org.junit.Assert.fail(Assert.java:88) 
     at org.junit.Assert.failNotEquals(Assert.java:743) 
     at org.junit.Assert.assertEquals(Assert.java:118) 
     at org.junit.Assert.assertEquals(Assert.java:555) 
     at LStringTest$LStringCompareToTest.t21aTestCompareTo(LStringTest.java:259) 
     ... 9 more 
6) t22aTestEquals[2](LStringTest$LStringCompareToTest) 
java.lang.AssertionError: equals of "a" and "ab" wrong expected:<false> but was:<true> 
     at org.junit.Assert.fail(Assert.java:88) 
     at org.junit.Assert.failNotEquals(Assert.java:743) 
     at org.junit.Assert.assertEquals(Assert.java:118) 
     at LStringTest$LStringCompareToTest.t22aTestEquals(LStringTest.java:269) 
     ... 9 more 
7) t21aTestCompareTo[3](LStringTest$LStringCompareToTest) 
java.lang.AssertionError: compareTo of "abc" and "abcd" wrong expected:<-1> but was:<0> 
     at org.junit.Assert.fail(Assert.java:88) 
     at org.junit.Assert.failNotEquals(Assert.java:743) 
     at org.junit.Assert.assertEquals(Assert.java:118) 
     at org.junit.Assert.assertEquals(Assert.java:555) 
     at LStringTest$LStringCompareToTest.t21aTestCompareTo(LStringTest.java:259) 
     ... 9 more 
8) t22aTestEquals[3](LStringTest$LStringCompareToTest) 
java.lang.AssertionError: equals of "abc" and "abcd" wrong expected:<false> but was:<true> 
     at org.junit.Assert.fail(Assert.java:88) 
     at org.junit.Assert.failNotEquals(Assert.java:743) 
     at org.junit.Assert.assertEquals(Assert.java:118) 
     at LStringTest$LStringCompareToTest.t22aTestEquals(LStringTest.java:269) 
     ... 9 more 
9) t21aTestCompareTo[4](LStringTest$LStringCompareToTest) 
java.lang.AssertionError: compareTo of "B" and "a" wrong expected:<-1> but was:<0> 
     at org.junit.Assert.fail(Assert.java:88) 
     at org.junit.Assert.failNotEquals(Assert.java:743) 
     at org.junit.Assert.assertEquals(Assert.java:118) 
     at org.junit.Assert.assertEquals(Assert.java:555) 

     at LStringTest$LStringCompareToTest.t21aTestCompareTo(LStringTest.java:259) 
     ... 9 more 
10) t22aTestEquals[4](LStringTest$LStringCompareToTest) 
java.lang.AssertionError: equals of "B" and "a" wrong expected:<false> but was:<true> 
     at org.junit.Assert.fail(Assert.java:88) 
     at org.junit.Assert.failNotEquals(Assert.java:743) 
     at org.junit.Assert.assertEquals(Assert.java:118) 
     at LStringTest$LStringCompareToTest.t22aTestEquals(LStringTest.java:269) 
     ... 9 more 
11) t21aTestCompareTo[5](LStringTest$LStringCompareToTest) 
java.lang.AssertionError: compareTo of "BB" and "Ba" wrong expected:<-1> but was:<0> 
     at org.junit.Assert.fail(Assert.java:88) 
     at org.junit.Assert.failNotEquals(Assert.java:743) 
     at org.junit.Assert.assertEquals(Assert.java:118) 
     at org.junit.Assert.assertEquals(Assert.java:555) 
     at LStringTest$LStringCompareToTest.t21aTestCompareTo(LStringTest.java:259) 
     ... 9 more 
12) t22aTestEquals[5](LStringTest$LStringCompareToTest) 
java.lang.AssertionError: equals of "BB" and "Ba" wrong expected:<false> but was:<true> 
     at org.junit.Assert.fail(Assert.java:88) 
     at org.junit.Assert.failNotEquals(Assert.java:743) 
     at org.junit.Assert.assertEquals(Assert.java:118) 
     at LStringTest$LStringCompareToTest.t22aTestEquals(LStringTest.java:269) 
     ... 9 more 

Test Failed! (12 of 18 tests failed.) 

Test failures: abandoning other phases.

的LString类是为了模仿Java的String和StringBuilder，但链表而不是阵列。我对如何使用this关键字有点困惑。在下面的compareTo()方法中，我想象使用this通过对自己说：“如果此索引处的此LStrings字符等于同一索引处的参数的LString字符，则返回0.”

我引用此页面，但不能确定如何将其有效地写：http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#compareTo%28java.lang.String%29

我想compareTo()返回0，如果LStrings有相同的字符，值小于零，如果这LString排在最后少而不是另一个字符串，如果按字典顺序更大，则其值大于零。

public class LString { 

    node front; 
    int size; 

    private class node { 
      char data; 
      node next; 

      public node(){ 
      } 

      public node (char newData){ 
       this.data = newData; 
      } 

      public node (char newData, node newNext){ 
       this.data = newData; 
       this.next = newNext; 
      } 


    } 

    public LString(){ 
      this.size = 0; 
      this.front = null; 
    } 
    public LString(String original) { 
     this.size = original.length(); 
     if (original != ""){ 

      this.front = new node(original.charAt(0)); 
      node curr = this.front; 

      for (int i =1; i < original.length(); i++) { 
       curr.next = new node(original.charAt(i)); 
       curr = curr.next; 
      } 
     } 



    } 


    public int length() { 
     return this.size; 
    } 
    public int compareTo(LString anotherLString) { 
     for (int i = 0; i < this.length(); i++) { 
     if (this.charAt(i) == anotherLString.charAt(i)) { 
      return 0; 
     } 
     //other logic for returning 1 or -1 
     } 
     return this.length()-anotherLString.length(); 
    } 
    public boolean equals(Object other) { 
     if (other == null || !(other instanceof LString)) { 
      return false; 
     } 
     else { 
      LString otherLString = (LString)other; 
      return true; 
     } 
    } 
    public char charAt(int index) { 
     //return this.charAt(index)? 
    } 
}

此代码中的方法未完成。任何意见，赞赏，试图学习Java。

来源

2015-03-03 rigatoni

你应该比较'数据'吗？ 'data'是一个单一的'char'？为什么不使用'String'或'char []'？ – Tdorno 2015-03-03 03:28:50

我在比较他们的词典值。 – rigatoni 2015-03-03 04:01:59

答案在http://stackoverflow.com/questions/28824474/lstring-class-using-linked-lists-to-make-strings-java/28826014#28826014中给出 – 2015-03-03 14:04:16

public int compareTo(LString anotherLString) { 
     for (int i = 0; i < this.length() && i<anotherLString.length(); i++) { 
      if (this.charAt(i) == anotherLString.charAt(i)) { 
       return 0; 
      } 
      else if(((int)this.charAt(i))<((int)anotherLString.charAt(i))) 
       return -1; 
      else 
       return 1; 
     } 
     return this.length()-anotherLString.length(); 
     }

我认为这个解决方案应该适合您的要求。我输入了char到int，然后比较它们以返回-1/1。此外，我还在for循环中添加了长度限制。

注：使用charAt(index)是不有效的解决方案在这里，因为你通过你的链接列表中的每个时间迭代你读取字符。相反，您应该直接遍历链表以进行比较。

来源

2015-03-04 06:08:17 Karthik

为链接列表中的字符编写compareTo（）方法

回答

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