[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?

解题思路

设链表长度为n，头结点与循环节点之间的长度为k。定义两个指针slow和fast，slow每次走一步，，fast每次走两步。当两个指针相遇时，有：

fast = slow * 2fast – slow = (n – k)的倍数 由上述两个式子可以得到slow为（n-k）的倍数

两个指针相遇后，slow指针回到头结点的位置，fast指针保持在相遇的节点。此时它们距离循环节点的距离都为k，然后以步长为1遍历链表，再次相遇点即为循环节点的位置。

实现代码/** * Definition for singly-linked list. * struct ListNode { *int val; *ListNode *next; *ListNode(int x) : val(x), next(NULL) {} * }; */ //Runtime:16 msclass Solution {public:ListNode *detectCycle(ListNode *head) {if (head == NULL){return NULL;}ListNode *slow = head;ListNode *fast = head;while (fast->next && fast->next->next){slow = slow->next;fast = fast->next->next;if (fast == slow){break;}}if (fast->next && fast->next->next){slow = head;while (slow != fast){slow = slow->next;fast = fast->next;}return slow;}return NULL;}};

那么，不如我们礼貌地保持相对距离，不至于太冷，不至于太痛。