我正在使用java中的链表，我需要获取x对象列表并将奇数位置的对象移动到列表的末尾。
我必须通过使用链接，没有新节点，没有list.data交换来实现。
当我将一个列表从一个列表移动到另一个列表时，我觉得我有一个不错的句柄，但是遍历和附加只引用一个列表是非常困难的。
这是实际问题---
编写一个方法移位，重新排列整数列表的元素，方法是移动到列表末尾所有奇数位置的值，否则保留列表顺序。例如，假设变量列表存储以下值：
[0,1,2,3,4,5,6,7] list.shift（）的调用;应该将列表重新排列为：
[0,2,4,6,1,3,5,7] 你必须通过重新安排列表的链接来解决这个问题。
下面是我之前需要编写该方法的类（具有上述限制。
我无法想出一个攻击计划。
// 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
}
答案 0 :(得分：1)
这样看：
首先我们需要某种光标，它将通过列表并指向我们的“当前”节点
第二，我们需要一些布尔变量（我将其称为INV）初始化为FALSE ...每次我们移动列表中的节点时，我们反转INV
如果你从左边开始查看列表，第二个元素是第一个要重新排列的元素，这将是我们的初始光标位置
让我们对该元素/节点进行引用，并将该引用保留为中止标准
循环开始：
现在从列表中删除当前节点并将其插入列表的末尾（移到结尾...而不是光标可能不随节点一起移动......）
将光标移动到我们刚移动的节点的前一个位置右侧的节点（如果存在）
如果当前元素是我们的中止标准（我们移动的第一个元素），我们可以假设列表现在按所需顺序排序 - ＆gt;我们完成了 - ＆gt;退出循环...如果不是我们的中止标准......继续
评估“光标的索引是偶数”为TRUE或FALSE ... XOR表示INV
如果结果为TRUE，则将光标移动到下一个元素...如果它为FALSE删除节点并将其插入到末尾（将其移至末尾）
做循环
-
这种方法在我们浏览列表时不会保留订单，但在完成后会按所需顺序排列...
INV var用于补偿当删除节点时索引移位...（0,1,2,3 ......如果你删除1并将其放在最后，2将有一个奇数索引，所以如果我们在每次移动时将其反转，我们就会得到“正确的”元素