• 142. Linked List Cycle II


    package LeetCode_142
    
    /**
     * 142. Linked List Cycle II
     * https://leetcode.com/problems/linked-list-cycle-ii/description/
     *
     * Given 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.
    Note that pos is not passed as a parameter.
    Notice that you should not modify the linked list.
    
    Follow up:
    Can you solve it using O(1) (i.e. constant) memory?
     * */
    
    class ListNode(var `val`: Int) {
        var next: ListNode? = null
    }
    
    class Solution {
        /*
        * solution: Two pointer, fast and slow
        * */
        fun detectCycle(head: ListNode?): ListNode? {
            if (head == null) {
                return null
            }
            //detect if had cycle
            var fast = head
            var slow = head
            var hasCycle = false
            //fast each take two steps
            //slow each take one step
            while (fast != null && fast.next != null) {
                fast = fast.next?.next
                slow = slow?.next
                if (fast == slow) {
                    hasCycle = true
                    break
                }
            }
            if (!hasCycle) {
                return null
            }
            //reset the fast
            fast = head
            //if fast and slow meet at a point, break
            while (fast != slow) {
                fast = fast?.next
                slow = slow?.next
            }
            return fast
        }
    }
  • 相关阅读:
    MySQL 元数据
    MySQL 复制表
    MySQL 临时表
    MySQL 索引
    MySQL ALTER
    MySQL 事务
    MySQL 正则表达式
    Mysql Join
    Python(数据库之表操作)
    Python知识点复习之__call__
  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/13670309.html
Copyright © 2020-2023  润新知