Copy List with Random Pointer——技巧
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.
分析:
我们知道如果是简单的copy List 的话,那么我们只需要从头到尾遍历下来,new出对应个数的Node,并把它们的连接关系设置好就可以了,但是这道题目中每个节点Node出现了Random属性,也就意味着可能当前结点Node所依赖的那个Random对应的结点还没有被创建出来。
如何能够找到新链表每个节点 random 域 所指向的节点呢??
思路: Step 1: 首先指向在原链表的每个节点后面,复制一个新的节点,原链表长度变为 2 倍
random 指针指向的是 原链表节点 random 指针指向的节点的后面的那个节点
Step 2: 将链表拆成两个 lists.
下面的代码不是自己写的,自己写的那个不知道为啥,超时了,也不知道错在哪里,这题还有一种用Map的解法,没去看,到时候好好想想········
/**
* 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) {
RandomListNode *tHead = head;
RandomListNode *next = NULL;
while(tHead)
{
next = tHead->next;
RandomListNode *node = new RandomListNode(tHead->label);
node->next = tHead->next;
//node->random = tHead->random;
tHead->next = node;
tHead= next;
}
tHead = head;
while(tHead)
{
if(tHead->random) tHead->next->random = tHead->random->next;
tHead = tHead->next->next;
}
RandomListNode *retHead = NULL;
RandomListNode *tRet = NULL; tHead = head;
RandomListNode *next2 = NULL;
while(tHead)
{
if(retHead == NULL)
{
next2 = tHead->next->next;
retHead = tHead->next;
tRet = retHead;
tHead->next = next2;
tHead = next2;
}
else
{
next2 = tHead->next->next;
tRet->next = tHead->next;
tHead->next = next2;
tHead = next2;
tRet = tRet->next;
} }
return retHead;
}
};
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