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) == 0:
            return None
        for i in range(len(l)-1,-1,-1):
            if i % 2 == 0:
                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