LeetCode 14 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.

分析：将复制的节点接在原来节点的后面，这样原来节点random所指节点的复制节点（即random.next）就是当前复制节点的random所要指向的节点。

/**
* 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 pointer = head;
//第一步，复制每个节点，接在该节点正后面
while(pointer != null){
RandomListNode copy = new RandomListNode(pointer.label);
copy.next = pointer.next;
pointer.next = copy;
pointer = pointer.next.next;
}
//第二步，处理random指针，被复制节点random指针所指节点的复制节点（即一下节点）
//便是复制节点random所应该指的地方
//注意处理random所指为null的情况
pointer = head;
while(pointer != null){
if(pointer.random == null){
pointer.next.random = null;
}else{
pointer.next.random = pointer.random.next;
}
pointer = pointer.next.next;
}
//第三步，要把两条链表解耦
//用node.next = node.next.next的方法即可解耦
//注意最后空指针的情况。
pointer = head;
RandomListNode newHead = head.next;
RandomListNode newPointer = newHead;
while(pointer != null){
pointer.next = pointer.next.next;
newPointer.next = (pointer.next==null) ? null : newPointer.next.next;

pointer = pointer.next;
newPointer = newPointer.next;
}
return newHead;
}
}