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;
    }
};