LeetCode -- Linked List Cycle II

题目描述：


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


Note: Do not modify the linked list.


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




判断链表是否有环，如果存在，返回环起始节点；如果不存在，返回Null。


思路：
1. 使用快慢指针的方法找到环的位置。
2. 如果找到了环，慢指针回到起点，快慢指针每次各走一步，下一次相遇的位置就是环的起点。




实现代码：

/** * Definition for singly-linked list. * public class ListNode { *     public int val; *     public ListNode next; *     public ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode DetectCycle(ListNode head)     {        if(head == null){    return null;    }        var p = head;    var q = head;        var found = false;    while(p != null && q != null && q.next != null && !found){    var t = q;    p = p.next;    q = q.next.next;    if(ReferenceEquals(p,q)){    found = true;    }    }            if(!found){    return null;    }        // p start from head again    // and q standing where it is    // next time they meet point is where cycle starts from    p = head;    while(!ReferenceEquals(p, q)){    p = p.next;    q = q.next;    }        return q;    }}