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复杂链表的复制
问题描写叙述:
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.
解题思路:
将1->2->3->4->NULL先复制为1->1->2->2->3->3->4->4->NULL。这样就会有例如以下关系式:
pCur->next->random = pCur->random->next;
/** * 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 *pCur = head, *tmp, *new_head; while(pCur != NULL)/*复制结点*/ { tmp = new RandomListNode(pCur->label); tmp->next = pCur->next; pCur->next = tmp; pCur = tmp->next; } pCur = head; /* 注意调用空指针的next指针。会报错*/ while(pCur != NULL)/*复制随意指针*/ { if (pCur->random != NULL) pCur->next->random = pCur->random->next; else pCur->next->random = NULL; pCur = pCur->next->next; } pCur = head; tmp = new_head = head->next; while(tmp != NULL)/* 将链表拆分为两个链表 */ { pCur->next = tmp->next; if(pCur->next == NULL) break; else tmp->next = pCur->next->next; pCur = pCur->next; tmp = tmp->next; } return new_head; } };
复杂链表的复制
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