[leetcode] Copy List with Random Pointer

Copy List with Random Pointer

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.

思路：

不会。看着提示显示的哈希表的题目，不清楚怎么用哈希表存储random的节点。看了别人的做法，豁然开朗。unordered_map存储的是本链表的某个节点和复制之后的链表对应的节点，即链表的第二个节点对应复制链表的第二个节点（深度复制，那当然两个节点一样）。第一步完成之后，链表的next域全部复制完成，同时节点也对应起来了。第二步，假设p1与p2为对应的节点，那么p1->random节点也会对应p2->random，所以又map[p1]->random = map[p1->random]。

题解：

/**
 * 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==NULL)
            return NULL;
        unordered_map<RandomListNode*, RandomListNode*> map;
        RandomListNode *head2 = new RandomListNode(head->label);
        RandomListNode *p1 = head, *p2 = head2;
        map[p1] = p2;
        p1 = p1->next;
        while(p1) {
            RandomListNode *tmp = new RandomListNode(p1->label);
            map[p1] = tmp;
            p2->next = tmp;
            p2 = tmp;
            p1 = p1->next;
        }
        p1 = head;
        while(p1) {
            if(p1->random)
                map[p1]->random = map[p1->random];
            p1 = p1->next;
        }
        return head2;
    }
};

后话：

还是参考了一篇博客。上面的方法空间复杂度为O(n)，还有一种O(1)的方法，感觉太牛逼了，看看就好，第一种方法要吃透，哈希表存储的新旧两个对应的节点很值得学习。

/**
 * 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==NULL)  return head;
        RandomListNode *pos1 = head, *pos2 = head->next;
        while(pos1!=NULL) {
            pos1->next = new RandomListNode(pos1->label);
            pos1->next->next = pos2;
            pos1 = pos2;
            if(pos2!=NULL)
                pos2 = pos2->next;
        }
        pos1 = head;
        pos2 = head->next;
        while(pos1!=NULL) {
            if(pos1->random==NULL)
                pos2->random==NULL;
            else
                pos2->random = pos1->random->next;
            pos1 = pos2->next;
            if(pos2->next!=NULL)
                pos2 = pos2->next->next;
        }
        RandomListNode *res = head->next;
        pos1 = head;
        pos2 = res;
        while(pos2->next!=NULL) {
            pos1->next = pos2->next;
            pos1 = pos2->next;
            pos2->next = pos1->next;
            pos2 = pos1->next;
        }
        pos1->next = NULL;
        pos2->next = NULL;
        return res;
    }
};