LeetCode #138 - Copy List with Random Pointer

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本文介绍两种深拷贝含随机指针链表的方法。第一种使用哈希表保存原链表和新链表中每个节点的对应关系，确保随机指针正确复制。第二种方法将新链表穿插在原链表中，实现O(1)空间复杂度的拷贝。

题目描述：

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.

/**
 * 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) {}
 * };
 */

// 用哈希表保存原链表和新链表中每个节点的对应关系，那么q->random = hash[p->random]
class Solution {
public:
    RandomListNode *copyRandomList(RandomListNode *head) {
        if (!head) return NULL;
        RandomListNode *res = new RandomListNode(head->label);
        RandomListNode *p = head->next;
        RandomListNode *q = res;
        unordered_map<RandomListNode*, RandomListNode*> hash;
        hash[head] = res;
        while (p) {//这个循环中p比q早一步，方便复制一个和p一样的节点，让q指向它
            RandomListNode *tmp = new RandomListNode(p->label);
            q->next = tmp;
            hash[p] = tmp;
            q = q->next;
            p = p->next;
        }

        //这个循环中p和q同步
        p = head;
        q = res;
        while (q) {
            q->random = hash[p->random];
            q = q->next;
            p = p->next;
        }
        return res;
    }
};

// 将新链表穿插在原链表中，空间复杂度为O(1)
class Solution {
public:
    RandomListNode *copyRandomList(RandomListNode *head)
    {
        if (!head) return NULL;
        RandomListNode *p = head;
        while (p)
        {
            RandomListNode *new_node = new RandomListNode(p->label);
            new_node->next = p->next;
            p->next = new_node;
            p = new_node->next;
        }
        p = head;
        while (p)
        {
            if (p->random) p->next->random = p->random->next; //复制random指针
            p = p->next->next; //一次跳两步
        }
        p = head;
        RandomListNode *res = head->next;
        while (p)
        {
            RandomListNode * q = p->next;
            if(p->next) p->next = p->next->next;
            if(q->next) q->next = q->next->next;
            p = p->next;
        }
        return res;
    }
};