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[LeetCode] Linked List Cycle II
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Follow up:
Can you solve it without using extra space?
思路：快慢指针的应用。时间复杂度O(n),空间复杂度O(1)
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution {10 public:11 ListNode *detectCycle(ListNode *head) {12 if (head == NULL) return head;13 ListNode *fast = head;14 ListNode *slow = head;15 16 while (fast != NULL && fast->next != NULL) {17 fast = fast->next->next;18 slow = slow->next;19 if (fast == slow) break;20 }21 22 if (fast == NULL || fast->next == NULL) 23 return NULL;24 25 slow = head;26 while (slow != fast) {27 slow = slow->next;28 fast = fast->next;29 }30 return slow;31 }32 };
[LeetCode] Linked List Cycle II
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