[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?

Linked List Cycle的扩展题，要找到cycle的起始点，还是跟Linked List Cycle那题一样，维护两个指针，fast指针以两倍slow指针的速度往前走 如果两个指针相遇，则有cycle  此时移动slow指针到head节点，这时俩指针离cycle的起点距离一样，两个指针以同样的速度往前走，就会在cycle的起点相遇~

这题的数学分析推导一直没耐心看 仅仅是记住了这种做法  找时间再好好看看

class Solution:
    # @param head, a ListNode
    # @return a list node
    def detectCycle(self, head):
        if head is None or head.next is None: return None
        slow, fast = head, head
        while fast != None and fast.next != None:
            fast, slow = fast.next.next, slow.next
            if fast == slow:
                break
        if fast != slow:
            return None
        slow = head
        while slow != fast:
            slow, fast = slow.next, fast.next
        return slow