LeetCode | 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.

Shallow copy:

Some members of the copy may reference the same objects as the original.

Deep copy:

All members of the original are cloned. There are no shared objects.

主要区别在于拷贝之后指针成员指向的是不是同一个地址。

这道题最naive的方法就是用一个unordered_map来存original pointer和copied pointer的对应关系。空间复杂度o(n),时间复杂度o(n)。

class Solution {
public:
    RandomListNode *copyRandomList(RandomListNode *head) {
        if (head == NULL) return NULL;
        unordered_map<RandomListNode*, RandomListNode*> pointerMap;
        
        RandomListNode *p = head, *copyH = NULL, *copyT = NULL, *tmp;
        while (p != NULL) {
            tmp = new RandomListNode(p->label);
            tmp->random = p->random;
            if (copyH == NULL) {
                copyH = tmp;
            } else {
                copyT->next = tmp;
            }
            copyT = tmp;
            pointerMap[p] = tmp;
            p = p->next;
        }
        
        p = copyH;
        
        while (p != NULL) {
            p->random = pointerMap[p->random];
            p = p->next;
        }
        
        return copyH;
    }
};

比较smart的方法是，将copied node放在对应的original node后面,也就是original node1->copied node1->original node2->copied node2->....

然后就可以用p->random->next来求得random的新地址了。 

最后再把copied list和original list分离开。

以后如果有题目需要把两个list对应起来的话，可以采用相同的方式，把它们串起来。

RandomListNode *copyRandomList(RandomListNode *head) {
    if(head == NULL) return NULL;
    RandomListNode *p = head;
    do {
        RandomListNode *q = p->next;
        p->next = new RandomListNode(p->label);
        p->next->next = q;
        p = q;
    } while(p != NULL);
    p = head;
    do {
        p->next->random = (p->random == NULL) ? NULL : p->random->next;
        p = p->next->next;
    } while(p != NULL);
    p = head;
    RandomListNode *r = head->next;
    for(RandomListNode *q = r;;) {
        p->next = q->next;
        p = p->next;
        if(p == NULL) break;
        q->next = p->next;
        q = q->next;
    }
    return r;
}