LeetCode: Linked List Cycle II

先用双指针探到是不是有回环，假设环长为D，head到环的距离（即要找的点到head的距离）为L，慢指针走的距离为L+X，快指针比慢指针快了一个或者多个环，所以为2*（L+X)，注意X < D，因为快指针比慢指针多走了一个或者多个环，所以L+X = k*D。这样解得X = k*D - L。这时离环头点距离为D-X = （1-k）*D+L。再来个节点从head开始走，都是慢指针，这样俩个指针会在环头处相遇。因为在环里的那个指针走过的距离为L，这样离环头点距离为L-（1-k）*D+L = （k-1）*D % D = 0。就是环头点。

/**
* 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 *p, *q;
p = q = head;
if (!head || !head->next) return NULL;
while (q && q->next) {
p = p->next;
q = q->next->next;
if (p == q) break;
}
if (p != q) return NULL;
p = head;
while (p != q) {
p = p->next;
q = q->next;
}
return p;
}
};

C#

/**
* Definition for singly-linked list.
* public class ListNode {
*     public int val;
*     public ListNode next;
*     public ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode DetectCycle(ListNode head) {
ListNode p = head, q = head;
if (head == null || head.next == null) return null;
while (q != null && q.next != null) {
p = p.next;
q = q.next.next;
if (p == q) break;
}
if (p != q) return null;
p = head;
while (p != q) {
p = p.next;
q = q.next;
}
return p;
}
}

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