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?

借用博客http://www.cnblogs.com/hiddenfox/p/3408931.html的图

设环的距离为L = (b+c)，无环的距离为a，

假设在时间t相遇，慢指针行驶距离为x，则 1 x t = x，即t=x

则快指针行驶的距离为2t = 2x，

则慢指针环上停留点为（x-a)%L，快指针停留点为 (2x-a)%L，由于在t时刻相遇，故（x-a)%L = (2x-a)%L，

根据同余定理 x%L = 0，

当快指针相遇后减慢速度为1，快指针从z继续行走a长度停下，则行走的路程为2x+a，在环上的位置为(2x+a-a)%L = 2x%L=2(x%L) = 0，即回到环的起始点，故知道环的起始点

ListNode* hasCycle(ListNode* head){
    if(head == NULL || head->next == NULL)  return false;
    ListNode* first = head, *second = head;
    while(second!=NULL && second->next!=NULL){
        first = first->next;
        second = second->next->next;
        if(first == second) return first;
    }
    return NULL;
}

ListNode *detectCycle(ListNode *head){
    ListNode* cycleNode = hasCycle(head);
    if(cycleNode != NULL){
        ListNode* startNode = head;
        while(startNode!=cycleNode){
            cycleNode = cycleNode->next;
            startNode = startNode->next;
        }
    }
    return cycleNode;
}