0142. Linked List Cycle I I

# 142. Linked List Cycle II#

## 题目 #

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.

## 解题思路 #

fast 和 slow 会相遇，说明他们走的时间是相同的，可以知道他们走的路程有以下的关系：

fast  t = (x1 + x2 + x3 + x2) / 2
slow  t = (x1 + x2) / 1

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

## 代码 #

package leetcode

/**
* Definition for singly-linked list.
* type ListNode struct {
*     Val int
*     Next *ListNode
* }
*/
func detectCycle(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return nil
}
isCycle, slow := hasCycle142(head)
if !isCycle {
return nil
}
for fast != slow {
fast = fast.Next
slow = slow.Next
}
return fast
}

func hasCycle142(head *ListNode) (bool, *ListNode) {
for slow != nil && fast != nil && fast.Next != nil {
fast = fast.Next.Next
slow = slow.Next
if fast == slow {
return true, slow
}
}
return false, nil
}

Sep 6, 2020