python3 解法, 执行用时: 44 ms, 内存消耗: 16.9 MB, 提交时间: 2022-11-27 11:06:52

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def oddEvenList(self, head: ListNode) -> ListNode:
        if not head:
            return head
        
        evenHead = head.next
        odd, even = head, evenHead
        while even and even.next:
            odd.next = even.next
            odd = odd.next
            even.next = odd.next
            even = even.next
        odd.next = evenHead
        return head

golang 解法, 执行用时: 0 ms, 内存消耗: 3.1 MB, 提交时间: 2022-11-27 11:06:36

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func oddEvenList(head *ListNode) *ListNode {
    if head == nil {
        return head
    }
    evenHead := head.Next
    odd := head
    even := evenHead
    for even != nil && even.Next != nil {
        odd.Next = even.Next
        odd = odd.Next
        even.Next = odd.Next
        even = even.Next
    }
    odd.Next = evenHead
    return head
}

javascript 解法, 执行用时: 68 ms, 内存消耗: 43.2 MB, 提交时间: 2022-11-27 11:06:21

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var oddEvenList = function(head) {
    if (head === null) {
        return head;
    }
    let evenHead = head.next;
    let odd = head, even = evenHead;
    while (even !== null && even.next !== null) {
        odd.next = even.next;
        odd = odd.next;
        even.next = odd.next;
        even = even.next;
    }
    odd.next = evenHead;
    return head;
};