题目
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.
思路
- 快慢指针的是否相遇，判断是否有环
- 判断环入口
当slow与fast相遇在环内某个节点，slow走了x个节点，fast走了2x个节点；
l为整个链表的长度，y为入口到slow的距离，z为head到入口的距离
2x = l+y
x = z+y
所以l-x = z, 也就是说这样新的节点与slow会在环开始的地方相遇
代码
package linkList;
/**
* Created by liqiushi on 2018/1/12.
*/
public class LinkedListCycleTwo {
public ListNode detectCycle(ListNode head) {
if(head == null||head.next == null){
return null;
}
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) {
ListNode ptr = head;
while (ptr != slow) {
ptr = ptr.next;
slow = slow.next;
}
return slow;
}
}
return null;
}
}
