[leetcode] 328. Odd Even Linked List @ python

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原题

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 …

解法1

将链表转化为list, 将列表的奇数和偶数元素分开, 然后组成新列表, 最后将列表转化为链表.
Time: O(n)
Space: O(n)

代码

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

class Solution:
    def oddEvenList(self, head: 'ListNode') -> 'ListNode':
        # get the list
        l = []
        while head:
            l.append(head.val)
            head = head.next
            
        l = l[::2] + l[1::2]        
        dummy = p = ListNode(0)        
        for val in l:
            p.next = ListNode(val)
            p = p.next
        return dummy.next

解法2

构造两个虚拟节点, 遍历head, 将odd连接奇数节点, even连接偶数节点, 退出循环后将两个链表连接起来, 最后返回新链表的头部节点.

Time: O(n)
Space: O(1)

代码

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

class Solution:
    def oddEvenList(self, head: 'ListNode') -> 'ListNode':
        d1 = odd = ListNode(0)        
        d2 = even = ListNode(0)
        while head:
            odd.next = head
            even.next = head.next
            odd = odd.next
            even = even.next
            head = head.next.next if even else None
        odd.next = d2.next
        return d1.next