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.
Take advantage of the original list, but there is a WA code. Don't break the structure after allocate the random pointers.
/**
* 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)
return NULL;
RandomListNode *cur = head, *curcopy = NULL, *next = NULL, *res =NULL, *dummy = new RandomListNode(-1), *l2 = dummy;
while (cur != NULL) {
next = cur->next;
cur->next = new RandomListNode(cur->label);
cur->next->next = next;
cur = next;
}
cur = head;
while (cur != NULL) {
next = cur->next->next;
l2->next = cur->next;
l2 = l2->next;
if (cur->random)
l2->random = cur->random->next;
cur->next = next;
cur = next;
}
res = dummy->next;
delete dummy;
return res;
}
};The right code is :
/**
* 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)
return NULL;
RandomListNode *cur = head, *curcopy = NULL, *next = NULL, *res =NULL, *dummy = new RandomListNode(-1), *l2 = dummy;
while (cur != NULL) {
next = cur->next;
cur->next = new RandomListNode(cur->label);
cur->next->next = next;
cur = next;
}
cur = head;
while (cur != NULL) {
if (cur->random)
cur->next->random = cur->random->next;
cur = cur->next->next;
}
cur = head;
while (cur != NULL) {
next = cur->next->next;
l2->next = cur->next;
l2 = l2->next;
cur->next = next;
cur = next;
}
res = dummy->next;
delete dummy;
return res;
}
};

本文介绍了一种复杂链表结构的深拷贝方法,该链表每个节点包含一个额外的随机指针。文章详细展示了如何通过一次遍历实现节点复制,并通过第二次遍历来设置随机指针,最后拆分原始链表和拷贝链表。
2023

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