我在一个名为equals(Object other)
的建筑物中遇到LString
方法时遇到问题。该类包含一个名为LString
的对象,该对象可以使用链接列表和其他一些方法构建字符串。我正在研究两种方法compareTo()
和equals()
。
通过运行另一个名为LStringTest
的文件来测试该类,该文件输出以下错误消息:
Running compareTo and equals tests (18 tests)
Starting tests: .E.........E...E.E
Time: 0.426
There were 4 failures:
1) t22aTestEquals[0](LStringTest$LStringCompareToTest)
java.lang.Exception: test timed out after 100 milliseconds
at LString.equals(LString.java:104)
at LStringTest$LStringCompareToTest.t22aTestEquals(LStringTest.java:269)
... 9 more
2) t22aTestEquals[5](LStringTest$LStringCompareToTest)
java.lang.Exception: test timed out after 100 milliseconds
at LString.equals(LString.java:104)
at LStringTest$LStringCompareToTest.t22aTestEquals(LStringTest.java:269)
... 9 more
3) t22aTestEquals[7](LStringTest$LStringCompareToTest)
java.lang.Exception: test timed out after 100 milliseconds
at LString.equals(LString.java:104)
at LStringTest$LStringCompareToTest.t22aTestEquals(LStringTest.java:269)
... 9 more
4) t22aTestEquals[8](LStringTest$LStringCompareToTest)
java.lang.Exception: test timed out after 100 milliseconds
at LString.equals(LString.java:104)
at LStringTest$LStringCompareToTest.t22aTestEquals(LStringTest.java:269)
... 9 more
Test Failed! (4 of 18 tests failed.)
Test failures: abandoning other phases.
我相信我compareTo()
和equals()
是正确的,我无法找到错误的根源。
这是我的代码:
import java.io.*;
import java.util.*;
public class LString {
node front;
int size;
//Creating a node class
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;
}
}
//Constructors
public LString(){
this.size = 0;
this.front = null;
}
public LString(String original) {
this.size = original.length();
if (original.length() > 0){
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;
}
}
}
// Length method, returns the length of LString
public int length() {
return this.size;
}
// compareTo method, compares this LString to anotherLString, returns 0 if equal,
// -1 if lexicogrpahically less, and 1 if lexicographically greater
public int compareTo(LString anotherLString) {
int len1 = length();
int len2 = anotherLString.length();
int lim = Math.min(len1, len2);
node cn1 = front;
node cn2 = anotherLString.front;
int k = 0;
while (k < lim) {
char c1 = cn1.data;
char c2 = cn2.data;
if (c1 != c2) {
return c1-c2;
}
k++;
cn1 = cn1.next;
cn2 = cn2.next;
}
return len1 - len2;
}
// a boolean equals method that returns true if LString and other are the same, false if not
public boolean equals(Object other) {
if (this == other) {
return true;
}
if (other instanceof LString) {
LString otherLString = (LString)other;
int n = length();
if (n == otherLString.length()) {
node n1 = front;
node n2 = otherLString.front;
while (n1 != null) {
if (n1.data != n2.data) {
return false;
}
}
return true;
}
}
return false;
}
}
感谢您提供任何建议,尝试学习java。
答案 0 :(得分:3)
您的equals
方法包含无限循环。你的while循环检查是否n1 != null
,但是在循环中没有更改n1
的值:
while (n1 != null) {
if (n1.data != n2.data) {
return false;
}
}
检查每个值后,您需要在列表中前进节点:
while (n1 != null) {
if (n1.data != n2.data) {
return false;
}
n1 = n1.next;
n2 = n2.next;
}