leetcode 之 Copy List with Random Pointer

题目如下：

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.

参考链接：http://www.cnblogs.com/TenosDoIt/p/3387000.html

题目大意：深拷贝一个链表，链表除了含有next指针外，还包含一个random指针，该指针指向字符串中的某个节点或者为空。

节点定义为：

struct RandomListNode {
int label;
RandomListNode *next, *random;
RandomListNode(int x) : label(x), next(NULL), random(NULL) {}
};

假设原始链表如下，细线表示next指针，粗线表示random指针，没有画出的指针均指向NULL：

算法1：我们在构建新链表的节点时，保存原始链表的next指针映射关系，并把指针做如下变化(蓝色为原始链表节点，紫红色为新链表节点)：

然后在上图的基础上进行如下两步

1、构建新链表的random指针：比如new1->random = new1->random->random->next, new2->random = NULL, new3-random = NULL, new4->random = new4->random->random->next

2、恢复原始链表：根据最开始保存的原始链表next指针映射关系恢复原始链表

该算法时间空间复杂度均为O(N)

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算法2：该算法更为巧妙，不用保存原始链表的映射关系，构建新节点时，指针做如下变化，即把新节点插入到相应的旧节点后面：

同理分两步

1、构建新节点random指针：new1->random = old1->random->next, new2-random = NULL, new3-random = NULL, new4->random = old4->random->next

2、恢复原始链表以及构建新链表：例如old1->next = old1->next->next,  new1->next = new1->next->next

该算法时间复杂度O(N)，空间复杂度O(1)

下面给出第二种算法的实现：

/** * 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) {        return recover(random(copy(head)));    }        private RandomListNode copy(RandomListNode head){        RandomListNode p = head;        while(p != null){            RandomListNode q = new RandomListNode(p.label);            q.random = null;            RandomListNode temp = p;            p = p.next;            q.next = p;            temp.next = q;        }        return head;    }        private RandomListNode random(RandomListNode head){        RandomListNode oldnode = head;        if(head == null) return null;        RandomListNode newnode = head.next;        while(oldnode != null){            if(oldnode.random != null)                newnode.random = oldnode.random.next;            oldnode = newnode.next;            if(oldnode != null)                newnode = oldnode.next;                }        return head;    }        private RandomListNode recover(RandomListNode head){        RandomListNode oldnode = head;        if(head == null) return null;        RandomListNode newHead = head.next;        RandomListNode newnode = newHead;        while(oldnode != null){            oldnode.next = newnode.next;            oldnode = newnode.next;            if(oldnode != null){                newnode.next = oldnode.next;                newnode = oldnode.next;            }        }        return newHead;    }}