Linked List Cycle II

Q:
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?

A:
就用CC150上的那道题目。  先slow and fast runner meet后，再把fast 放到开头，重新跑（速度变慢）



Mistakes:
1： update fastRunner 到链表头的时候，忘了把slowRunner也往下走一个。

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        if(head == null || head.next == null)
            return null;
        ListNode slow = head, fast = head.next;
        while(slow!= fast){
            slow = slow.next;
            fast = fast.next;
            if(fast==null || fast.next == null)
                return null;
            fast = fast.next;
        }
        slow = head;
        fast = fast.next;
        while(slow!= fast){
            slow = slow.next;
            fast = fast.next;
        }
        return slow;
    }
}

证明如下：
假设前面不重复的链表长度为k,     # of nodes in circle is n.
at time t = k+a   (where a is the number of nodes slow have run in the circle)
=======>  a+k = 2(k+a) - x*n ,  (x倍的n), 取代了求模运算
则有 a+k = x*n
那么，fast再走 k时间，也就是再走 (a +k) 的时候，就回到了原点。
这里有一个注意的地方就是，（当我们重新计数的时候，slow=head时，已经表明我们走了一步了。因此我们需要把fast也前进一步。


Learned: