leetcode刷题（142）——判断单链表是否有环，如果有环，输出环起始位置索引

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给定一个链表，返回环的开始节点。如果链表无环，则返回null。介绍了如何判断链表中是否存在环，并在存在环的情况下找出环的起始位置。解决方案涉及到快慢指针的概念，通过它们移动距离的关系来确定环的起始点。

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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: Do not modify the linked list.

Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.
Follow up:
Can you solve it without using extra space?

To understand this solution, you just need to ask yourself these question.
Assume the distance from head to the start of the loop is x1
the distance from the start of the loop to the point fast and slow meet is x2
the distance from the point fast and slow meet to the start of the loop is x3
What is the distance fast moved? What is the distance slow moved? And their relationship?

x1 + x2 + x3 + x2
x1 + x2
x1 + x2 + x3 + x2 = 2 (x1 + x2)

Thus x1 = x3

Finally got the idea.

/**
 * 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) {
        ListNode slow = head;
        ListNode fast = head;
        while(slow!=null&&fast!=null&&fast.next!=null){
            slow = slow.next;
            fast = fast.next.next;
            if(fast==slow){
                ListNode slow2 = head;
                while(slow2!=slow){
                    slow = slow.next;
                    slow2 = slow2.next;
                }
                return slow;
            }
        }
        return null;
    }
}