假设有如下一个链表:
1
2
3
4
5
6
|
struct
Node { int
value ; struct
Node *next ; struct
Node *random ; } |
1
|
Node
*deepCopy (Node *head) |
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)
代码:Node *deepCopy (Node *head) { Node* now = head; Node* next = head->next; while( now != NULL ) { Node * copy = new Node; copy->value = now->value; copy->next = now->next; now ->next = copy; now = next; next = next->next; } now = head; while( now != NULL ) { now->next->random = now->random->next; now = now->next->next; } Node* head2 = head->next; Node* newHead = head->next; while( head2->next != NULL ) { head->next = head2->next; head2->next = head2->next->next; head = head->next; head2 = head2->next; } return newHead; }
|
版权声明:本文为博主原创文章,未经博主允许不得转载。
创新工场笔试题Copy List with Random Pointer
原文地址:http://blog.csdn.net/u014082714/article/details/46953891