【leetcode】328. Odd Even Linked List

题目如下：

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:

• 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 ...

解题思路：本题难度不大，分别维护奇/偶节点的两个头指针，然后遍历链表，分别将指针指向对方的next，指向后再把自己移动到之前指向的next即可。

代码如下：

# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution(object):
    def oddEvenList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head == None or head.next == None:
            return head
        evenHead = even = head.next
        odd = head
        while odd != None and even != None:
            odd.next = even.next
            if odd.next != None:
                odd = odd.next
            even.next = odd.next
            even = even.next
        odd.next = evenHead
        return head

 

转载于:https://www.cnblogs.com/seyjs/p/10193058.html