138. Copy List with Random Pointer（unsolved）

2019-11-06 07:08:00

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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： step1: 复制l1链表中的每个节点分别连接到该节点自身的后面。如1->1->2->2->4->4->3->3->null

step2:每一个原节点后面的那个节点，即拷贝的节点，复制原节点的random的next，因为是要复制整条链表。

step3:拆分两条链表。分别对l1,l2操作

/** * 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) { RandomListNode *newHead, *l1, *l2; if (head == NULL) return NULL; for (l1 = head; l1 != NULL; l1 = l1->next->next) { l2 = new RandomListNode(l1->label); l2->next = l1->next; l1->next = l2; } newHead = head->next; for (l1 = head; l1 != NULL; l1 = l1->next->next) { if (l1->random != NULL) l1->next->random = l1->random->next; } for (l1 = head; l1 != NULL; l1 = l1->next) { l2 = l1->next; l1->next = l2->next; if (l2->next != NULL) l2->next = l2->next->next; } return newHead; }};

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