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