问题描述:
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.
问题分析:本题转换了一下思路,将这个链表看作一颗二叉树,对其进行先根遍历构造新树(新的链表)。
过程详见代码:
/**
* 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*> m;
return bl(m, head);
}
RandomListNode * bl(unordered_map<RandomListNode*, RandomListNode*>& m, RandomListNode *head)
{
if (head == NULL) return NULL;
if (!m.count(head))
{
RandomListNode * root = new RandomListNode(head->label);
m[head] = root;
root->next = bl(m, head->next);
root->random = bl(m, head->random);
return root;
}
else return m[head];
}
};