Description
Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.
There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail’s next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.
Do not modify the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1 Output: tail connects to node index 1 Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0 Output: tail connects to node index 0 Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1 Output: no cycle Explanation: There is no cycle in the linked list.
Constraints:
- The number of the nodes in the list is in the range
[0, 104]. -105 <= Node.val <= 105posis-1or a valid index in the linked-list.
Follow up: Can you solve it using O(1) (i.e. constant) memory?
Code
Time Complexity: , Space Complexity:
一脈相承 Linked List Cycle。找 Cycle 起點可參考 Floyd’s Cycle Finding Algorithm。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode *detectCycle(struct ListNode *head) {
struct ListNode* slow = head;
struct ListNode* fast = head;
while(slow && fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
if(slow == fast) {
slow = head;
while(slow != fast) {
slow = slow->next;
fast = fast->next;
}
return slow;
}
}
return NULL;
}/**
* 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;
bool flag = false;
while(slow && fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
if(slow == fast) {
flag = true;
break;
}
}
if(!flag) return nullptr;
slow = head;
while(slow != fast) {
slow = slow->next;
fast = fast->next;
}
return slow;
}
};