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.
- 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 ...
Example:
- Given
1->2->3->4->5->NULL, - return
1->3->5->2->4->NULL.
C Solution:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* oddEvenList(struct ListNode* head) {
struct ListNode odd, even, *o = &odd, *e = &even;
o->next = 0;
e->next = 0;
int isodd = 1;
while (head) {
if (isodd) {
o->next = head;
o = o->next;
}
else {
e->next = head;
e = e->next;
}
isodd ^= 1;
head = head->next;
}
o->next = even.next;
e->next = 0;
return odd.next;
}
Python Solution:
# 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
"""
ocur = odd = ListNode(0)
ecur = even = ListNode(0)
is_ocur = 1
while head:
if is_ocur:
ocur.next = ocur = head
else:
ecur.next = ecur = head
head = head.next
is_ocur ^= 1
ecur.next = None
ocur.next = even.next
return odd.next
Summary:
- 6ms, 8.41%
- When it's necessary to record the head and the tail, use dummy and its pointer is suitable.
LeetCode: 328. Odd Even Linked List
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