Determine whether a linked list has a cycle, and return the node where the cycle begins. If there is no cycle, return NULL.
Detecting whether a linked list has a cycle without allocating extra space was mentioned in a previous blog post:
http://blog.sina.com.cn/s/blog_672f71fc0101odsf.html
How to find the entry node of the cycle?
If we can calculate the number of nodes k in the cycle, then using the approach of removing the nth node from the end:
http://blog.sina.com.cn/s/blog_672f71fc0101pfx6.html
This problem can be solved.
Here is the code:
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| class Solution {
public:
ListNode *detectCycle(ListNode *head) {
if(head == NULL)
return NULL;
ListNode * p1 = head;
ListNode * p2 = head->next;
if(p2 == NULL)
return NULL;
p2 = head->next;
while(true)
{
if(p2 == NULL)
{
return NULL;
}
else
if(p1 == p2)
{
break;
}
else
{
p1 = p1->next;
p2 = p2->next;
if(p2 == NULL)return NULL;
p2 = p2->next;
}
}
int k = 1;
p1 = p1->next;
p2 = p2->next->next;
while(p1 != p2)
{
p1 = p1->next;
p2 = p2->next->next;
k++;
}
p1 = head;
p2 = head;
for(int i = 0;i < k;i++)
{
p2 = p2->next;
}
while(p1 != p2)
{
p1 = p1->next;
p2 = p2->next;
}
return p1;
}
};
|