[LeetCode] 328. Odd Even Linked List

题：https://leetcode.com/problems/odd-even-linked-list/description/

题目

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 1:

Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL

Example 2:

Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL

Note:

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

思路

题目大意

一个 link，将link 从1 开始编号，将link重新排序，将编号为奇数的node放到 前面，偶数的放到后面。

解题思路

方法一 设置奇、偶link

将原link分解，生成 奇数link 与 偶数link。
然后将 偶数link 加在奇数link的尾部。

方法二 奇偶指针穿插

在原link上，设置 oddp，evenp直接在link上修改。

code

方法一 设置奇、偶link

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        ListNode oddlink = new ListNode(0);
        ListNode evenlink = new ListNode(0);
        
        ListNode  oddp = oddlink;
        ListNode  evenp = evenlink;
        
        int i = 1;
        while(head!=null){
            ListNode p = head;
            head = head.next;
            p.next = null;
            if(i %2 == 1){
                oddp.next = p;
                oddp = oddp.next;

            }
            else{
                evenp.next = p;
                evenp = evenp.next;
            }
            i++;
        }
        oddp.next = evenlink.next;
        return oddlink.next;

    }
}

方法二 奇偶指针穿插

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode oddEvenList(ListNode head) {
        if(head == null || head.next == null)   return    head;
        ListNode  oddp = head;
        ListNode  evenp = head.next;
        ListNode evenhead = evenp;
        while(evenp != null && evenp.next != null){
            oddp.next = evenp.next;
            oddp = oddp.next;
            evenp.next = oddp.next;
            evenp = evenp.next;
        }
        oddp.next = evenhead;
        return head;

    }
}