[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。设fast为快指针，每次步进两个节点，慢指针为slow，每次步进一个节点。设环长度为Y，环入口距离起点的距离为X,。两个指针在环内距离入口的第k个节点相遇。当两个指针相遇时，慢指针共走了 S步（S=X+nY+K )，快指针共走了2S步（2S= X+mY+K）步。所以

2×（X+nY+K) = X+mY+K 

=> (m-2n)Y = X+K

=> X = (m-2n-1)Y+(Y-K)

所以另慢指针从起点开始走，每次步进一个节点。快指针从相遇的节点开始每次步进一个节点。则两个指针最终会在环入口处相遇。

如果m-2n-1=0，那么Y-K=X，快指针直接达到入口处。如果不为零，那么快指针将要在圈内转上几个来回才能在入口处与满指针相遇。

无论上述哪个情况，两个指针都走了X步。

时间复杂度O(n),空间复杂度O(1)

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