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java - 在java中交错两个节点列表

转载 作者:行者123 更新时间:2023-12-01 13:26:31 25 4
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我发布了一个类似的问题,但仍然遇到问题,尝试交错两个节点列表时遇到问题,根据我的测试用例,我应该得到 null,但它给出了一个包含“3D”的节点。

当我在 Eclipse 的 Debug模式下查看列表时,列表的持续时间似乎比应有的时间要长。测试了我的长度方法,它返回正确的长度。

我不明白为什么我的节点似乎会继续运行,而它应该只是下面发布的测试用例中的 13 个节点的列表。

代码应执行以下操作例如 {"KH","4C","8C","QC","3D","7D","JD"} 变为 {"KH","4C", "8C","QC"} {"3D","7D","JD"}(如果在本例中列表具有奇数个节点,则第一个列表更长)并返回列表 {"KH","3D","4C","7D","8C","JD","QC"}。

我必须使用提供的节点类来解决这个问题,并且不能使用静态变量、数组或 Java 集合。请不要只是发布代码,我真的需要学习如何操作节点,但不明白我所做的事情是错误的。

public class ListShuffleExample {

public static Node<String> shuffle(Node<String> deck) {
if (deck == null) {
return null;
}
if (deck.next == null) {
return deck;
}

Node<String> lf = deck;
int decklength = length(lf);
int halflength;
Node<String> first = deck;
Node<String> second = deck;
if (decklength % 2 == 0) {
halflength = decklength / 2;

} else {
halflength = (decklength / 2) + 1;
}

for (int i = 0; i < halflength; i++) {
second = second.next;
}

Node<String> curFirst = first;
Node<String> curSecond = second;

while (curFirst != null && curSecond != null) {
// save the next one of the first list
Node<String> nextFirst = curFirst.next;

// set the next one of the first list to the first one of the second
// list
curFirst.next = curSecond;

// save the next one of the second list
Node<String> nextSecond = curSecond.next;

// set the next one after the inserted item to the previous next of
// the first list
curSecond.next = nextFirst;

// set the current references to the next ones
curFirst = nextFirst;
curSecond = nextSecond;
}

return first;
}

public static int length(Node<String> adeck) {
int length = 0;

while (adeck != null) {
length++;
adeck = adeck.next;
}

return length;
}
}

测试用例

import static org.junit.Assert.*;

import java.lang.reflect.Field;

import org.junit.Test;

public class ListShuffleExampleTest {
private static final Node<String> _AH = new Node<String>( "AH" );
private static final Node<String> _5H = new Node<String>( "5H" );
private static final Node<String> _9H = new Node<String>( "9H" );
private static final Node<String> _KH = new Node<String>( "KH" );
private static final Node<String> _4C = new Node<String>( "4C" );
private static final Node<String> _8C = new Node<String>( "8C" );
private static final Node<String> _QC = new Node<String>( "QC" );
private static final Node<String> _3D = new Node<String>( "3D" );
private static final Node<String> _7D = new Node<String>( "7D" );
private static final Node<String> _JD = new Node<String>( "JD" );
private static final Node<String> _2S = new Node<String>( "2S" );
private static final Node<String> _6S = new Node<String>( "6S" );
private static final Node<String> _TS = new Node<String>( "TS" );

private static <T> Node<T> makeList(Node<T>... nodes) {
for (int i = 0; i < nodes.length-1; i++) {
nodes[ i ].next = nodes[ i+1 ];
}
nodes[ nodes.length-1 ].next = null;
return nodes[ 0 ];
}
@Test
public void testReflection() {
Class<?> iClass = ListShuffleExample.class;
for (Field field : iClass.getDeclaredFields()) {
if (!field.isSynthetic()) {
fail( "class should not have any fields" );
}
}
}
@Test
public void testNull() {
Node<String> actual = ListShuffleExample.shuffle( null );
//expected: NULL
assertEquals( "", null, actual );
}
@Test
public void testOddDivide10() {
@SuppressWarnings("unchecked")
Node<String> input = makeList( _QC );
Node<String> actual = ListShuffleExample.shuffle( input );
for (Object expected : new Object[]{ _QC }) {
assertEquals( "Incorrect value", expected, actual );
actual = actual.next;
}
assertNull( "Incorrect result", actual );
}
@Test
public void testOddDivide21() {
@SuppressWarnings("unchecked")
Node<String> input = makeList( _AH, _5H, _9H );
Node<String> actual = ListShuffleExample.shuffle( input );
// _AH, _5H
// _9H
for (Object expected : new Object[]{ _AH, _9H, _5H }) {
assertEquals( "Incorrect value", expected, actual );
actual = actual.next;
}
assertNull( "Incorrect result", actual );
}
@Test
public void testEvenDivide22() {
@SuppressWarnings("unchecked")
Node<String> input = makeList( _2S, _6S, _TS, _AH );
Node<String> actual = ListShuffleExample.shuffle( input );
// _2S, _6S
// _TS, _AH
for (Object expected : new Object[]{ _2S, _TS, _6S, _AH }) {
assertEquals( "Incorrect value", expected, actual );
actual = actual.next;
}
assertNull( "Incorrect result", actual );
}
@Test
public void testOddDivide43() {
@SuppressWarnings("unchecked")
Node<String> input = makeList( _KH, _4C, _8C, _QC, _3D, _7D, _JD );
Node<String> actual = ListShuffleExample.shuffle( input );
// _KH, _4C, _8C, _QC,
// _3D, _7D, _JD
for (Object expected : new Object[]{ _KH, _3D, _4C, _7D, _8C, _JD , _QC }) {
assertEquals( "Incorrect value", expected, actual );
actual = actual.next;
}
assertNull( "Incorrect result", actual );
}
@Test
public void testEvenDivide44() {
@SuppressWarnings("unchecked")
Node<String> input = makeList( _5H, _9H, _KH, _4C, _8C, _QC, _3D, _7D );
Node<String> actual = ListShuffleExample.shuffle( input );
// _5H, _9H, _KH, _4C,
// _8C, _QC, _3D, _7D
for (Object expected : new Object[]{ _5H, _8C, _9H, _QC, _KH, _3D, _4C, _7D }) {
assertEquals( "Incorrect value", expected, actual );
actual = actual.next;
}
assertNull( "Incorrect result", actual );
}
@Test
public void testMany() {
@SuppressWarnings("unchecked")
Node<String> input = makeList( _AH, _5H, _9H, _KH, _4C, _8C, _QC, _3D, _7D, _JD, _2S, _6S, _TS );
Node<String> actual = ListShuffleExample.shuffle( input );
// _AH, _5H, _9H, _KH, _4C, _8C, _QC
// _3D, _7D, _JD, _2S, _6S, _TS
for (Object expected : new Object[]{ _AH, _3D, _5H, _7D, _9H, _JD, _KH, _2S, _4C, _6S, _8C, _TS, _QC }) {
assertEquals( "Incorrect value", expected, actual );
actual = actual.next;
}
assertNull( "Incorrect result", actual );
}
}

节点类

public final class Node<T> {
public final T value;
public Node<T> next;

public Node(T _value) {
this( _value, null );
}
public Node(T _value, Node<T> _next) {
value = _value;
next = _next;
}
@Override
public String toString() {
return "" + value;
}
}

最佳答案

您似乎忘记取消第一个列表末尾与第二个列表开头的链接。

您希望列表的末尾指向 null,但您却留下了一个返回到第二个列表的原始开头的链接。当您有偶数分区时,它会起作用,因为最后一个元素已经 null 终止。但是,当您有奇数分区时,您需要考虑额外的元素。

编辑:循环可以通过两个条件的组合终止:curFirst == nullcurSecond == null 或两者兼而有之。您应该考虑是什么导致循环终止,以及哪些条件必须为真。另外,考虑是否还有剩余元素。

(您特意要求我们不要发布代码!)

关于java - 在java中交错两个节点列表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21791220/

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