Linked List Cycle II 找链表环入口 @LeetCode

经典链表环的问题

package Level3;

import Utility.ListNode;

/**
* Linked List Cycle II
*
*  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?
*
*/
public class S130 {

public static void main(String[] args) {

}

public ListNode detectCycle(ListNode head) {
if(head == null || head.next ==null){
return null;
}
ListNode slow = head;
ListNode fast = head;

// 找到第一次相遇点
while(fast!=null && fast.next!=null && slow!=null){
slow = slow.next;
fast = fast.next.next;
if(slow == fast){		// 相遇
break;
}
}

// 检查是否因为有环退出或是因为碰到null而退出
if(slow==null || fast==null || fast.next==null){
return null;
}

// 把慢指针移到链表头，然后保持快慢指针同样的速度移动
// 再次相遇时即为环的入口
slow = head;
while(slow != fast){
slow = slow.next;
fast = fast.next;
}

// 现在快慢指针都指向环的入口
return slow;
}

}

重写时的注意点写在注解里了

/**
* 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; // No node or single node ==> No cycle
}
ListNode fast = head;
ListNode slow = head;
while(fast!=null && fast.next!=null && slow!=null){
fast = fast.next.next;
slow = slow.next;
if(fast == slow){
break;
}
}

if(fast != slow){ // no cycle
return null;
}

slow = head;
while(fast!=null && slow!=null){
if(fast == slow){ // Check first! eg: 1,2
break;
}
fast = fast.next;
slow = slow.next;
}
return fast;
}
}