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;
    }
}