leetcode[138] Copy List with Random Pointer

这里是复制带有一个random指针的链表。是不是很熟悉啊。之前有做过克隆无向图的。那就借助leetcode Clone Graph的思路。

分两次遍历链表，一次先复制普通的含next的，另一次就是复制random了。利用map记录，可以一次就找到想要的点。

/**

 * 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)

    {

        if (!head) return head;

        RandomListNode *pre = new RandomListNode(0), *proc = head, *root = pre, *tmp_pre = pre;

        unordered_map<RandomListNode *, RandomListNode*> umap;

        while (proc)

        {

            root -> next = new RandomListNode(proc -> label);

            umap[proc] = root -> next;

            proc = proc -> next;

            root = root -> next;

        }

        while (head)

        {

            tmp_pre -> next -> random = umap[head -> random];

            head = head -> next;

            tmp_pre = tmp_pre -> next;

        }

        return pre -> next;

    }

};