gpt4 book ai didi

java - 为链表对象编写 equals() 方法

转载 作者:行者123 更新时间:2023-12-01 22:25:34 25 4
gpt4 key购买 nike

我在使用我正在构建的名为 LString 的类中的 equals(Object other) 方法时遇到问题。该类包含一个名为 LString 的对象,该对象从链接列表中构建字符串,以及一些其他方法。我正在研究两种方法,compareTo()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.

我相信我的 compareTo()equals() 是正确的,但我找不到错误的根源。

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;
}
}

感谢您的建议,正在努力学习java。

最佳答案

您的equals方法包含无限循环。您的 while 循环检查是否 n1 != null,但在循环中没有任何地方更改 n1 的值:

    while (n1 != null) {
if (n1.data != n2.data) {
return false;
}
}

检查每个值后,您需要将节点在列表中向下推进:

while (n1 != null) {
if (n1.data != n2.data) {
return false;
}
n1 = n1.next;
n2 = n2.next;
}

关于java - 为链表对象编写 equals() 方法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28844026/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com