Total Accepted: 53943 Total Submissions: 209664 Difficulty: Hard
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.
o(n)空间复杂度,代码比价简单。网上还流传一种o(1)空间复杂度的解法,大致的过程就是先拷贝next结点,再连接随机结点,最后分离链表。
/** * 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) { unordered_map<RandomListNode*,RandomListNode*> umap; RandomListNode* newHead = NULL; RandomListNode* cur = head; RandomListNode* node_pre = NULL; RandomListNode* node = NULL; while(cur){ node = new RandomListNode(cur->label); umap[cur] = node; cur == head ? newHead = node :node_pre->next = node; node_pre = node; cur = cur->next; } cur = head; while(cur){ umap[cur]->random = cur->random ? umap[cur->random] : NULL; cur = cur->next; } return newHead; } };
Next challenges: (M) Clone Graph
时间: 2024-10-22 15:20:38