描述
给定一个链表，如果链表中存在环，则返回到链表中环的起始节点的值，如果没有环，返回 null。
样例
给出 -21->10->4->5, tail connects to node index 1，返回 10
代码实现
/**
* Definition for ListNode.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int val) {
* this.val = val;
* this.next = null;
* }
* }
*/
public class Solution {
/**
* @param head: The first node of linked list.
* @return: The node where the cycle begins.
* if there is no cycle, return null
*/
public ListNode detectCycle(ListNode head) {
if (head == null || head.next == null) {
return null;
}
ListNode fast = head.next;
ListNode slow = head;
while (fast != slow) {
if (fast == null || fast.next == null) {
return null;
}
fast = fast.next.next;
slow = slow.next;
}
/**while (head != fast.next) {
head = head.next;
slow = fast.next;
}**/
//用慢指针的next与head比较
while (head != slow.next) {
head = head.next;
slow = slow.next;
}
return head;
}
}
