leetcode fast slow pointer

Introduction

In questions related to linked list, fast and slow pointer solution is very common. Fast pointer step two and Slow pointer step one. Always we can the regular through draw a picture and mathematical derivation, we can find:

• middle of the linked list
• interaction pointer to assert if there is a cycle

leetcode 141 Linked List Cycle

Question

Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

Solution

Only if there is a cycle, fast will catch up with the slow pointer.

class Solution(object):    def hasCycle(self, head):        """        :type head: ListNode        :rtype: bool        """        fast = head        slow = head        while fast != None and fast.next != None:            fast = fast.next.next            slow = slow.next            if fast == slow:                return True        return False

leetcode 142 Linked List Cycle II

Question

Fast pointer catch up slow when slow is on the first cycle. so:

• slowLen = a + b
• fastLen = (a + b) + n*r
• fastLen = 2 * slowLen
• a + b + n*r = a +b +b +c + (n-1)r = 2 (a + b)
• c = a +( n - 1) * r

When they come across, slow from the head and fast from the meeting point, they will finally meet at the begin point of the cycle.

class Solution(object):    def detectCycle(self, head):        """        :type head: ListNode        :rtype: ListNode        """        fast = head        slow = head        while fast and fast.next:            fast = fast.next.next            slow = slow.next            if fast == slow:                slow = head                while fast != slow:                    fast = fast.next                    slow = slow.next                return slow        return None