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我正在使用java中的链接列表，我需要获取x个对象的列表并将奇数位置的对象移动到列表的末尾。java list通过重新排列链接来移动项目
我必须通过使用链接，没有新的节点，没有list.data交换。
当我从一个列表移动到另一个列表时，我觉得我有一个体面的句柄，但遍历和追加只有一个列表的引用是非常艰难的。
这里的实际问题---
编写通过移动到列表中的为奇数位置否则保留列表顺序的所有值结束重新排列整数列表中的元素的方法转移。例如，假设变量列表存储以下值：
[0,1,2,3,4,5,6,7] list.shift（）;应重新排列列表为：
[0,2,4,6,1,3,5,7] 您必须通过重新排列列表的链接来解决此问题。
下面
是，我需要编写前的方法（与上述限制类。
我真的不能拿出一个行动的计划。
// A LinkedIntList object can be used to store a list of integers.
public class LinkedIntList {
private ListNode front; // node holding first value in list (null if empty)
private String name = "front"; // string to print for front of list
// Constructs an empty list.
public LinkedIntList() {
front = null;
}
// Constructs a list containing the given elements.
// For quick initialization via Practice-It test cases.
public LinkedIntList(int... elements) {
this("front", elements);
}
public LinkedIntList(String name, int... elements) {
this.name = name;
if (elements.length > 0) {
front = new ListNode(elements[0]);
ListNode current = front;
for (int i = 1; i < elements.length; i++) {
current.next = new ListNode(elements[i]);
current = current.next;
}
}
}
// Constructs a list containing the given front node.
// For quick initialization via Practice-It ListNode test cases.
private LinkedIntList(String name, ListNode front) {
this.name = name;
this.front = front;
}
// Appends the given value to the end of the list.
public void add(int value) {
if (front == null) {
front = new ListNode(value, front);
} else {
ListNode current = front;
while (current.next != null) {
current = current.next;
}
current.next = new ListNode(value);
}
}
// Inserts the given value at the given index in the list.
// Precondition: 0 <= index <= size
public void add(int index, int value) {
if (index == 0) {
front = new ListNode(value, front);
} else {
ListNode current = front;
for (int i = 0; i < index - 1; i++) {
current = current.next;
}
current.next = new ListNode(value, current.next);
}
}
public boolean equals(Object o) {
if (o instanceof LinkedIntList) {
LinkedIntList other = (LinkedIntList) o;
return toString().equals(other.toString()); // hackish
} else {
return false;
}
}
// Returns the integer at the given index in the list.
// Precondition: 0 <= index < size
public int get(int index) {
ListNode current = front;
for (int i = 0; i < index; i++) {
current = current.next;
}
return current.data;
}
// Removes the value at the given index from the list.
// Precondition: 0 <= index < size
public void remove(int index) {
if (index == 0) {
front = front.next;
} else {
ListNode current = front;
for (int i = 0; i < index - 1; i++) {
current = current.next;
}
current.next = current.next.next;
}
}
// Returns the number of elements in the list.
public int size() {
int count = 0;
ListNode current = front;
while (current != null) {
count++;
current = current.next;
}
return count;
}
// Returns a text representation of the list, giving
// indications as to the nodes and link structure of the list.
// Detects student bugs where the student has inserted a cycle
// into the list.
public String toFormattedString() {
ListNode.clearCycleData();
String result = this.name;
ListNode current = front;
boolean cycle = false;
while (current != null) {
result += " -> [" + current.data + "]";
if (current.cycle) {
result += " (cycle!)";
cycle = true;
break;
}
current = current.__gotoNext();
}
if (!cycle) {
result += " /";
}
return result;
}
// Returns a text representation of the list.
public String toString() {
return toFormattedString();
}
// Returns a shorter, more "java.util.LinkedList"-like text representation of the list.
public String toStringShort() {
ListNode.clearCycleData();
String result = "[";
ListNode current = front;
boolean cycle = false;
while (current != null) {
if (result.length() > 1) {
result += ", ";
}
result += current.data;
if (current.cycle) {
result += " (cycle!)";
cycle = true;
break;
}
current = current.__gotoNext();
}
if (!cycle) {
result += "]";
}
return result;
}
// ListNode is a class for storing a single node of a linked list. This
// node class is for a list of integer values.
// Most of the icky code is related to the task of figuring out
// if the student has accidentally created a cycle by pointing a later part of the list back to an earlier part.
public static class ListNode {
private static final List<ListNode> ALL_NODES = new ArrayList<ListNode>();
public static void clearCycleData() {
for (ListNode node : ALL_NODES) {
node.visited = false;
node.cycle = false;
}
}
public int data; // data stored in this node
public ListNode next; // link to next node in the list
public boolean visited; // has this node been seen yet?
public boolean cycle; // is there a cycle at this node?
// post: constructs a node with data 0 and null link
public ListNode() {
this(0, null);
}
// post: constructs a node with given data and null link
public ListNode(int data) {
this(data, null);
}
// post: constructs a node with given data and given link
public ListNode(int data, ListNode next) {
ALL_NODES.add(this);
this.data = data;
this.next = next;
this.visited = false;
this.cycle = false;
}
public ListNode __gotoNext() {
return __gotoNext(true);
}
public ListNode __gotoNext(boolean checkForCycle) {
if (checkForCycle) {
visited = true;
if (next != null) {
if (next.visited) {
// throw new IllegalStateException("cycle detected in list");
next.cycle = true;
}
next.visited = true;
}
}
return next;
}
}
// YOUR CODE GOES HERE
}
在纸上绘制这样的结构通常是一个很好的准备步骤来开始攻击... – 2012-08-02 15:21:48
也许从你的源列表中首先创建两个列表'0 2 4 6'和'1 3 5 7'，然后将它们链接起来最后。 – 2012-08-02 15:23:47
我认为这是'[作业]'？ – 2012-08-02 15:24:01