leetcode328题 题解 翻译 C语言版 Python版

328. Odd Even Linked List

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 ...

328.奇偶链表

给定一个单链表，把所有的奇数结点连在一起然后后面连上所有的偶数结点。请注意这里我们所说的奇偶是指结点编号而不是说结点里的值。

你应该尝试就地完成。程序应该是O(1)的空间复杂度和O(结点数)的时间复杂度。

举例：

给定 1->2->3->4->5->NULL

返回1->3->5->2->4->NULL

注意：

奇数结点组和偶数结点组内的结点相对顺序应该保持原来的相对顺序。

第一个结点是奇数结点，第二个结点是偶数结点，依次类推。

思路：遍历一次单链表即可，将奇数结点连在一起，偶数结点连在一起，最后在把偶数结点构成的链表添加到奇数结点构成的链表末尾。记录奇数链表可以直接用给定的原链表表头指针head，记录偶数结点链表可以新建一个指针even，然后设置两个游动指针，同时向右游动，将相连的一奇一偶结点改变他的下家结点。注意如果原链表本身为空要单独处理，因为前面的方法会访问head->next，当head为NULL时会报错。

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     struct ListNode *next;
* };
*/
struct ListNode* oddEvenList(struct ListNode* head) {
if (!head) return head;
struct ListNode *even, *p, *q;
even = head->next;
p = head;
q = even;
while (q && q->next){
p->next = q->next;
p = p->next;
q->next = p->next;
q = q->next;
}
p->next = even;
return head;
}

# 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 not head: return head
even = head.next
p = head
q = even
while q and q.next:
p.next = q.next
p = p.next
q.next = p.next
q = q.next
p.next = even
return head