leetcode fast slow pointer-优快云博客

本文详细介绍了快慢指针法在解决链表问题中的应用，包括判断链表是否有环、寻找环的入口点等。通过数学推导和代码实现，深入浅出地解释了这一经典算法。

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

cycle
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

cycle
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