leetcode-day12:Linked List Cycle II(环形链表II)

leetcode-day12:Linked List Cycle II(环形链表II) 142. linked List Cycle II

链表：环找到了，那入口呢？
Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail’s next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

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.

Constraints:

The number of the nodes in the list is in the range [0, 104].
-105 <= Node.val <= 105
pos is -1 or a valid index in the linked-list.

Follow up: Can you solve it using O(1) (i.e. constant) memory?

JAVA实现代码：
代码逻辑：

• 1.是否有环
  定义快慢指针，慢指针每次移动1个节点，快指针每次移动2个节点，相遇则有环
• 2.环入口在哪儿
  
  • 2.1 slow走过了x + y，fast走过了 x + y + n (y + z)，n为fast指针在环内走的圈数
  • 2.2 sow每次走1个节点，fast每次走2个节点，因此(x + y) * 2 = x + y + n (y + z)

  x + y = n (y + z)                           =》
  x = n (y + z) - y                           =》
  x = (n - 1) (y + z) + z                     =》
  n>=1,假设n=1=>x=z
  

  • 2.3 即：当快慢指针相遇时，慢指针继续从y走，从头节点重新开始一个节点向后，这俩节点将相遇在入口处。

public class Solution {
    public ListNode detectCycle(ListNode head) {
       ListNode slow = head;
		ListNode fast = head;
		
		boolean isHave=false;
		while(fast!=null && fast.next!=null){
			slow=slow.next;
			fast=fast.next.next;
            if(slow==fast){
				isHave=true;
				break;
			}
		}
		
		if(!isHave){
			return null;
		}
		
		ListNode slowHead = head;
		
		while(true){
            if(fast==slowHead){
				break;
			}
			fast=fast.next;
			slowHead=slowHead.next;

		}
		
		return slowHead;
    }
}