Copy List with Random Pointer

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.

解析：

含有random 指针的链表复杂问题，由于在剑指offer上看过类似的题，所以很快就有解题思路。

1,，先复杂链表将复制节点插入到原节点的后面

2，复制random指针，将复制节点的random指针指向原节点的random的next，即指向了random的复制节点

3.拆分链表，奇数节点构成了原链表，偶数节点构成了复制结果链表

很容易写出代码：

/** * Definition for singly-linked list with a random pointer. * struct RandomListNode { *     int label; *     RandomListNode *next, *random; *     RandomListNode(int x) : label(x), next(NULL), random(NULL) {} * }; */class Solution {public:    RandomListNode *copyRandomList(RandomListNode *head) {        if(head == NULL) return NULL;                RandomListNode *copyhead;        RandomListNode *ptemp = head, *pnew = NULL;        while(ptemp != NULL) //复制next         {        pnew = new RandomListNode(ptemp->label);        pnew->next = ptemp->next;        ptemp->next = pnew;        ptemp = pnew->next;        }        ptemp = head;RandomListNode *pcopyhead = head->next, *pcopy = head->next; while(ptemp != NULL) //复制random {    if( ptemp->random != NULL)pcopy->random = ptemp->random->next;ptemp = ptemp->next->next;if(pcopy->next == NULL) break;pcopy = pcopy->next->next;}ptemp = head;pcopyhead = pcopy = head->next;while(ptemp != NULL){ptemp->next = ptemp->next->next;if(pcopy->next == NULL) break;pcopy->next = pcopy->next->next;ptemp = ptemp->next;pcopy = pcopy->next;}return pcopyhead;    }};

很多时候思路比努力更重要，选择对的方向才能更好的有解决方法。fight