【原创】leetCodeOj --- Copy List with Random Pointer 解题报告

题目地址：

https://oj.leetcode.com/problems/copy-list-with-random-pointer/

题目内容：

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.

/**
* 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) {}
* };

*/

方法：

难点在于random指针需要指向一个复制过程中还不存在的结点。解决办法就简单了，先复制完单链表，再处理每个结点的random指针，这样复制random指针时其指向的结点就已经存在了。

为了高效完成这个工作，我们需要建立一个映射，以旧地址为键，新地址为值，方便复制random指针。

全部代码：

class Solution {
public:
    RandomListNode *copyRandomList(RandomListNode *head) {
        if (!head)
            return NULL;
        unordered_map<int,int> dict;
        RandomListNode *tmpHead = head;
        RandomListNode *resHead = NULL;
        RandomListNode *now     = NULL;
        RandomListNode *nxt     = NULL;
        resHead = (RandomListNode *)malloc(sizeof(RandomListNode));
        dict[(int)head] = (int)resHead;
        resHead->label = tmpHead->label;
        resHead->next  = NULL;
        resHead->random = NULL;
        now  = resHead;
        tmpHead = tmpHead->next;
        while (tmpHead)
        {
            nxt = (RandomListNode *)malloc(sizeof(RandomListNode));
            nxt->label = tmpHead->label;
            nxt->next  = NULL;
            nxt->random = NULL;
            dict[(int)tmpHead] = (int)nxt;
            now->next = nxt;
            now = nxt;
            tmpHead = tmpHead->next;
        }
        tmpHead = head;
        while (tmpHead)
        {
            now = (RandomListNode *)dict[(int)tmpHead];
            if (tmpHead->random != NULL)
                now->random = (RandomListNode *)dict[(int)(tmpHead->random)];

            tmpHead = tmpHead->next;
        }
        return resHead;
    }
};