leetcode--138. Copy List with Random Pointer

1，问题描述

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.

数据结构：

/**
 * Definition for singly-linked list with a random pointer.
 * class RandomListNode {
 *     int label;
 *     RandomListNode next, random;
 *     RandomListNode(int x) { this.label = x; }
 * };
 */

2，边界条件：root==null情况可以处理，所以没有边界条件，不需要特殊处理

3，解题思路：先不管random指针，把链表节点依次复制，不过不是生成一个新的链表，而是把新节点放在原节点的next。然后再根据random指针把新节点的random指针指向原节点的next，即新节点。最后把大链表分离，恢复旧链表，把新节点生成一个新链表，即为旧链表的deep copy。

4，代码实现

public RandomListNode copyRandomList(RandomListNode head) {
    if (head == null) {
        return null;
    }
    RandomListNode cur = head;
    while (cur != null) {
        RandomListNode node = new RandomListNode(cur.label);
        node.next = cur.next;
        cur.next = node;
        cur = node.next;
    }

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

    RandomListNode dummy = new RandomListNode(-1);
    // RandomListNode copyCur = head.next;//这种写法在leetcode不通过，在lintcode上面能通过
    // dummy.next = copyCur;
    // cur = head.next.next;
    RandomListNode copyCur = dummy;//这里是注意下，比上面写法简洁，可以处理head=null的情况
    cur = head;
    while (cur != null && cur.next != null) {
        copyCur.next = cur.next;
        cur.next = cur.next.next;
        cur = cur.next;
        copyCur = copyCur.next;
    }
    return dummy.next;
}

5，时间复杂度：O（n），空间复杂度：O（1）

6，api：无