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;
}