328. Odd Even Linked List

abson 發(fā)布于2019-07-31 10:19 / 509人閱讀

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.

https://leetcode.com/problems...

# 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:
        # If zero, one or two elements, then solved
        if head == None or head.next == None or head.next.next == None:
            return head

        # Two pointers
        p = head
        n = head.next
        t = None

        while n.next:
            # If there is even element
            if n.next.next:
                # Keep it for now for both n and p
                t = n.next.next
                m = p.next
                p.next = n.next
                p = p.next
                # Recover link for p and n
                p.next = m
                n.next = t
                n = n.next
            else:
                # Save and insert odd
                t = p.next
                p.next = n.next
                p = p.next
                p.next = t
                n.next = None

        return head

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