【python3】leetcode 328. Odd Even Linked List（Medium）

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328. Odd Even Linked List（Medium）

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 ...

 下标为偶数的排在下标为奇数的后面，下标按 奇，偶，奇，偶，....排列

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

class Solution:
    def oddEvenList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        A = []
        cp = head
        while(cp):
            A.append(cp)
            cp = cp.next
        if len(A) <= 2:return head
        for i in range(len(A)-2):
                A[i].next = A[i+2]
        if len(A) % 2 == 0:
            A[-2].next = A[1]
            A[-1].next = None
        else:
            A[-1].next = A[1]
            A[-2].next = None
        return head