leetcode328

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode OddEvenList(ListNode head) {
        if (head != null)
            {
                ListNode odd = head, even = head.next, evenHead = even;

                while (even != null && even.next != null)
                {
                    odd.next = odd.next.next;
                    even.next = even.next.next;
                    odd = odd.next;
                    even = even.next;
                }
                odd.next = evenHead;
            }
            return head;
    }
}

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

补充一个python的实现：

 class Solution:
     def oddEvenList(self, head: 'ListNode') -> 'ListNode':
         l = list()
         while head:
             l.append(head.val)
             head = head.next

         even = None
         odd = None
         temp = None
         if len(l) == :
             return None
         for i in range(len(l)-,-,-):
             if i %  == :
                 curodd = ListNode(l[i])
                 if not temp:
                     temp = curodd
                 curodd.next = odd
                 odd = curodd
             else:
                 cureven = ListNode(l[i])
                 cureven.next = even
                 even = cureven
         temp.next = even
         return odd

补充一个简化版本的，执行效率更高：

 class Solution:
     def oddEvenList(self, head: ListNode) -> ListNode:
         if head == None:
             return None
         odd,even,evenhead = head,head.next,head.next
         while even != None and even.next != None:
             odd.next = odd.next.next
             odd = odd.next
             even.next = even.next.next
             even = even.next
         odd.next = evenhead
         return head