linked-list-cycle-ii

Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull.

Follow up:
Can you solve it without using extra space?



/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public static ListNode detectCycle(ListNode head) {
        if (null == head || null == head.next) {
            return null;
        }
        if (head.next == head) {
            return head;
        }
        ListNode tempNode = head.next;
        head.next = null;
        ListNode result = detectCycle(tempNode);
        head.next = tempNode;
        if (null == result) {
            ListNode currentPtr = head.next;
            while (null != currentPtr) {
                if (currentPtr == head) {
                    result = head;
                    break;
                }
                currentPtr = currentPtr.next;
            }
        }
        return result;
    }
}

Others:

[@wangxiaobao](https://www.nowcoder.com/profile/122270)
思路：
1）同linked-list-cycle-i一题，使用快慢指针方法，判定是否存在环，并记录两指针相遇位置(Z)；
2）将两指针分别放在链表头(X)和相遇位置(Z)，并改为相同速度推进，则两指针在环开始位置相遇(Y)。

证明如下：
如下图所示，X,Y,Z分别为链表起始位置，环开始位置和两指针相遇位置，则根据快指针速度为慢指针速度的两倍，可以得出：
2*(a + b) = a + b + n * (b + c)；即
a=(n - 1) * b + n * c = (n - 1)(b + c) +c;
注意到b+c恰好为环的长度，故可以推出，如将此时两指针分别放在起始位置和相遇位置，并以相同速度前进，当一个指针走完距离a时，另一个指针恰好走出 绕环n-1圈加上c的距离。
故两指针会在环开始位置相遇。

![](http://uploadfiles.nowcoder.net/images/20150812/122270_1439340467801_QQ%E6%88%AA%E5%9B%BE20150812084712.jpg)

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if(head == NULL){
            return 0;
        }
        ListNode* slow = head;
        ListNode* fast = head;
        while(fast != NULL && fast->next != NULL){
            slow = slow->next;
            fast = fast->next->next;
            if(slow == fast){
                break;
            }
        }
        if(fast == NULL || fast->next == NULL){
            return NULL;
        }
        slow = head;
        while(slow != fast){
            slow = slow->next;
            fast = fast->next;
        }
        return slow;
    }
};
