[leetcode]Linked List Cycle II

http://oj.leetcode.com/problems/linked-list-cycle-ii/

老题。当快慢指针相交时，通过方程或观察可知，从head到环开始点的距离和从相遇点开始是一样的，那么从相遇点和从开始点再一起走直到相遇就行了。

/**
 * 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 *fast = head;
        ListNode *slow = head;
        do {
            if (fast == NULL || fast->next == NULL) return NULL;
            fast = fast->next->next;
            slow = slow->next;
        } while (fast != slow);
        ListNode *n1 = head;
        ListNode *n2 = slow;
        while (n1 != n2) {
            n1 = n1->next;
            n2 = n2->next;
        }
        return n1;
    }
};