leetcode328
2024-09-04 18:08:26
/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode OddEvenList(ListNode head) {
if (head != null)
{
ListNode odd = head, even = head.next, evenHead = even; while (even != null && even.next != null)
{
odd.next = odd.next.next;
even.next = even.next.next;
odd = odd.next;
even = even.next;
}
odd.next = evenHead;
}
return head;
}
}
https://leetcode.com/problems/odd-even-linked-list/#/description
补充一个python的实现:
class Solution:
def oddEvenList(self, head: 'ListNode') -> 'ListNode':
l = list()
while head:
l.append(head.val)
head = head.next even = None
odd = None
temp = None
if len(l) == :
return None
for i in range(len(l)-,-,-):
if i % == :
curodd = ListNode(l[i])
if not temp:
temp = curodd
curodd.next = odd
odd = curodd
else:
cureven = ListNode(l[i])
cureven.next = even
even = cureven
temp.next = even
return odd
补充一个简化版本的,执行效率更高:
class Solution:
def oddEvenList(self, head: ListNode) -> ListNode:
if head == None:
return None
odd,even,evenhead = head,head.next,head.next
while even != None and even.next != None:
odd.next = odd.next.next
odd = odd.next
even.next = even.next.next
even = even.next
odd.next = evenhead
return head
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