Linked List Cycle II
Desicription
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Note: Do not modify the linked list.
Follow up:
Can you solve it without using extra space?
Solution
代码语言：javascript复制/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* detectCycle(ListNode* head) {
ListNode* slow = head;
ListNode* fast = head;
ListNode* entry = head;
while(fast && fast->next && fast->next->next) {
slow = slow->next;
fast = fast->next->next;
if(slow == fast) {
while(slow != entry) {
slow = slow->next;
entry = entry->next;
}
return entry;
}
}
return NULL;
}
};
