Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up: Can you solve it without using extra space? Observation: Solution: /** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode detectCycle(ListNode head) { if (head == null) return null; ListNode slow = head; ListNode fast = head; boolean hasCycle = false; while (slow.

by lek tin in "algorithm" access_time 2-min read

Linked List Cycle

Given a linked list, determine if it has a cycle in it. Follow-up Can you solve it without using extra space? Solution # Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def hasCycle(self, head): """ :type head: ListNode :rtype: bool """ slow = head fast = head while fast: # fast reaches the end, no cycle was detected if fast and not fast.

by lek tin in "algorithm" access_time 1-min read

Remove Nth Node From End of List

Given a linked list, remove the n-th node from the end of list and return its head. Example Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5. Note Given n will always be valid. Follow-up Could you do this in one pass? Solution # Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.

by lek tin in "algorithm" access_time 1-min read