【leetcode】No.138：copy-list-with-

题目描述

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.

思路：

先新建copy节点插入到每一个节点后，然后遍历复制random指针，最后将两个链表拆分开，注意插入后步长为2，细心点就好。

代码：

/**
 * 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;
        // 添加新节点到旧节点后面
        RandomListNode cur = head;
        while (cur != null){
            insertNode(cur);
            cur = cur.next.next;
        }
        // 设置random指针
        cur = head;
        while (cur != null){
            if (cur.random != null){
                cur.next.random = cur.random.next;
            }
            else {
                cur.next.random = null;
            }
            cur = cur.next.next;
        }
        // 取下新节点
        cur = head;
        RandomListNode newHead = head.next;
        while (cur != null){
            divideNode(cur);
            cur = cur.next;
        }
        return newHead;
    }

    private void insertNode(RandomListNode cur){
        RandomListNode newNode = new RandomListNode(cur.label);
        newNode.next = cur.next;
        cur.next = newNode;
    }

    private void divideNode(RandomListNode cur){
        RandomListNode newCur = cur.next;
        cur.next = newCur.next;
        if (newCur.next != null){
            newCur.next = newCur.next.next;
        }
    }
}