[LeetCode]Odd Even Linked List

题目描述:

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:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->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 ...

题目大意:

给定一个单链表,将其节点进行分组,使得所有的奇数节点排列在前,偶数节点在后。请注意这里的奇偶指的是节点序号而不是节点的值。

你应当尝试“就地”完成此问题。程序应当满足O(1)的空间复杂度和O(nodes)的时间复杂度。

测试用例见题目描述。

注意:

偶数与奇数节点分组内部的相对顺序应当与输入保持一致。

第一个节点为奇数节点,第二个节点为偶数节点,以此类推。

解题思路:

链表的基本操作,详见代码。

Python代码:

# 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 is None: return head
        odd = oddHead = head
        even = evenHead = head.next
        while even and even.next:
            odd.next = even.next
            odd = odd.next
            even.next = odd.next
            even = even.next
        odd.next = evenHead
        return oddHead

 

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