Linked List Cycle II
描述
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?
题解
/**
* 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 *fast = head, *slow = head, *entry = head;
while(fast != nullptr && fast->next != nullptr) {
slow = slow->next;
fast = fast->next->next;
if (fast == slow) {
while(slow != entry) {
slow = slow->next;
entry = entry->next;
}
return entry;
}
}
return NULL;
}
};