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Two methods for doing this problem:
1. With extra memory, using hashtable:
I made a mistake for mapping[runner->random] = mapping[runner]->random. Then it will erase original one and not correctly copied.
1 /** 2 * Definition for singly-linked list with a random pointer. 3 * struct RandomListNode { 4 * int label; 5 * RandomListNode *next, *random; 6 * RandomListNode(int x) : label(x), next(NULL), random(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 RandomListNode *copyRandomList(RandomListNode *head) { 12 if (!head) return NULL; 13 RandomListNode *result = new RandomListNode(head->label); 14 unordered_map<RandomListNode *, RandomListNode *> mapping; 15 mapping[head] = result; 16 RandomListNode *runner = head; 17 while (runner->next) { 18 if (mapping.find(runner->next) == mapping.end()) { 19 RandomListNode *newNode = new RandomListNode(runner->next->label); 20 mapping[runner->next] = newNode; 21 mapping[runner]->next = newNode; 22 } 23 runner = runner->next; 24 } 25 runner = head; 26 while (runner) { 27 mapping[runner]->random = mapping[runner->random]; 28 runner = runner->next; 29 } 30 return result; 31 } 32 };
2. Merge two list into one and then split them:
A stupid mistake:
when last split two list, I use
while (head->next)
It will cause problem. Because when your head go to the end, the head is NULL, then head->next is an illegal argument.
1 /** 2 * Definition for singly-linked list with a random pointer. 3 * struct RandomListNode { 4 * int label; 5 * RandomListNode *next, *random; 6 * RandomListNode(int x) : label(x), next(NULL), random(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 RandomListNode *copyRandomList(RandomListNode *head) { 12 if (!head) return NULL; 13 RandomListNode *runner = head; 14 RandomListNode *result = new RandomListNode(0); 15 while (runner) { 16 RandomListNode *newNode = new RandomListNode(runner->label); 17 newNode->next = runner->next; 18 runner->next = newNode; 19 runner = runner->next->next; 20 } 21 runner = head; 22 while (runner) { 23 if (runner->random != NULL) { 24 runner->next->random = runner->random->next; 25 } 26 runner = runner->next->next; 27 } 28 runner = result; 29 while (head) { 30 runner->next = head->next; 31 head->next = head->next->next; 32 head = head->next; 33 runner = runner->next; 34 } 35 return result->next; 36 } 37 };
LeetCode – Refresh – Copy List with Random Pointer
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原文地址：http://www.cnblogs.com/shuashuashua/p/4349357.html
