• 328. Odd Even Linked List


    Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

    You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

    Example:
    Given 1->2->3->4->5->NULL,
    return 1->3->5->2->4->NULL.

    Note:
    The relative order inside both the even and odd groups should remain as it was in the input. 
    The first node is considered odd, the second node even and so on ...

    Credits:
    Special thanks to @DjangoUnchained for adding this problem and creating all test cases.

    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
     * }
     */
    public class Solution {
        public ListNode oddEvenList(ListNode head) {
            if(head == null || head.next == null || head.next.next == null) return head;
            ListNode odd = head;
            ListNode even = head.next;
            ListNode secondNode = even;
            while(even != null && even.next != null){
                odd.next = even.next;
                odd = odd.next;
                even.next = odd.next;
                even = even.next;
            }
            odd.next = secondNode;
            return head;
        }
    }
  • 相关阅读:
    闰年or平年判断
    输入一个日期判断是否正确的几种方法
    网页布局+下拉隐藏栏
    360导航布局
    [LeetCode] Longest Common Prefix
    [LeetCode] Length of Last Word
    [LeetCode] Valid Palindrome II
    [Qt] Qt信号槽
    [LeetCode] Split Linked List in Parts
    [LeetCode] Find Pivot Index
  • 原文地址:https://www.cnblogs.com/joannacode/p/6128379.html
Copyright © 2020-2023  润新知