Copy List with Random Pointer 复制有随机指针的链表@LeetCode

1 先画好图
2 注意可能形成环的情况，因此必须拆成3个循环来做

代码是按照 http://www.cnblogs.com/lautsie/p/3259724.html 的图来写的
一个单链表，其中除了next指针外，还有一个random指针，指向链表中的任意某个元素。如何复制这样一个链表呢？通过next来复制一条链是很容易的，问题的难点在于如何恰当地设置新链表中的random指针。
很容易想到使用Hash表的做法，先依次遍历原链表，每经过一个节点X，开辟一个新节点Y，然后（key=X的地址，value=Y的地址）存入哈希表。第二次再遍历原链表，根据拓扑结构设置新的链表。需要O(n)的空间，时间也是O(n)。如果不使用额外的空间，那么要想在旧链表和新链表的对应节点之间建立联系。就要利用链表中多余的指针。
O(n)复杂度，O(1)空间。如图所示，扫描两边即可。需要复制的链表:

如图所示，ABCD是原来的链表，A’B’C’D’是复制的链表，第一遍扫描顺序复制next指针，把ABCD的next分别指向A’B’C’D’，将A’的next指针指向B，B’的next指针指向C，依次类推：

复制random指针： A’->random=A->random->next恢复:A->next=A’->next;A’->next=A’->next->next;

再来一张示例图：




http://www.cnblogs.com/reynold-lei/p/3362614.html

package Level4;

import Utility.RandomListNode;

/**
* Copy List with Random Pointer
*
* A linked list is given such that each node contains an additional random
* pointer which could point to any node in the list or null.
*
* Return a deep copy of the list.
*
*/
public class S138 {

public static void main(String[] args) {
RandomListNode head = new RandomListNode(1);
RandomListNode n2 = new RandomListNode(2);
RandomListNode n3 = new RandomListNode(3);
RandomListNode n4 = new RandomListNode(4);
head.next = n2;
n2.next = n3;
n3.next = n4;
head.random = n3;
n2.random = n4;
RandomListNode ret = copyRandomList(head);
while(ret != null){
System.out.println(ret.label);
ret = ret.next;
}
}

public static RandomListNode copyRandomList(RandomListNode head) {
if(head == null){
return null;
}

RandomListNode cur = head;
RandomListNode copyHead = null;
RandomListNode copyCur = null;

// 形成图1
while(cur != null){
copyCur = new RandomListNode(cur.label);
if(copyHead == null){
copyHead = copyCur;
}
copyCur.next = cur.next;
cur.next = copyCur;
cur = cur.next.next;
}

// 形成图2
// 注意此处必须要分成两个while loop而不能合并。
// 如果合并，则在两对random指针互相形成环时会出错！
cur = head;
while(cur != null){
cur.next.random = cur.random!=null ? cur.random.next : null; // 设置copy节点的random指针
cur = cur.next.next;
}

cur = head;
while(cur != null){
copyCur = cur.next;
cur.next = cur.next.next; // 恢复cur节点的next指针
copyCur.next = cur.next!=null ? cur.next.next : null; // 设置copy节点的next指针
cur = cur.next;
}

return copyHead;
}
}


记住图和三个步骤，然后细心
/**
* Definition for singly-linked list with a random pointer.
* class RandomListNode {
* int label;
* RandomListNode next, random;
* RandomListNode(int x) { this.label = x; }
* };
*/
public class Solution {
public RandomListNode copyRandomList(RandomListNode head) {
if(head == null) {
return null;
}
RandomListNode copyHead = null;
RandomListNode copyCur = null;
RandomListNode cur = head;

while(cur != null) {
copyCur = new RandomListNode(cur.label);
if(copyHead == null) {
copyHead = copyCur;
}
copyCur.next = cur.next;
cur.next = copyCur;
cur = cur.next.next;
}

cur = head;
copyCur = copyHead;
while(cur != null) {
if(cur.random != null) {
copyCur.random = cur.random.next;
}
cur = cur.next.next;
if(copyCur.next != null) {
copyCur = copyCur.next.next;
}
}

cur = head;
copyCur = copyHead;
while(cur != null) {
cur.next = cur.next.next;
if(copyCur.next != null) {
copyCur.next = copyCur.next.next;
}
cur = cur.next;
copyCur = copyCur.next;
}

return copyHead;
}
}