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;
    }
}