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

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