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我正在使用java中的鏈接列表，我需要獲取x個對象的列表並將奇數位置的對象移動到列表的末尾。java list通過重新排列鏈接來移動項目
我必須通過使用鏈接，沒有新的節點，沒有list.data交換。
當我從一個列表移動到另一個列表時，我覺得我有一個體面的句柄，但遍歷和追加只有一個列表的引用是非常艱難的。
這裏的實際問題---
編寫通過移動到列表中的爲奇數位置否則保留列表順序的所有值結束重新排列整數列表中的元素的方法轉移。例如，假設變量列表存儲以下值：
[0,1,2,3,4,5,6,7] list.shift（）;應重新排列列表爲：
[0,2,4,6,1,3,5,7] 您必須通過重新排列列表的鏈接來解決此問題。
下面
是，我需要編寫前的方法（與上述限制類。
我真的不能拿出一個行動的計劃。
// 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
}
在紙上繪製這樣的結構通常是一個很好的準備步驟來開始攻擊... – 2012-08-02 15:21:48
也許從你的源列表中首先創建兩個列表'0 2 4 6'和'1 3 5 7'，然後將它們鏈接起來最後。 – 2012-08-02 15:23:47
我認爲這是'[作業]'？ – 2012-08-02 15:24:01