python3 解法, 执行用时: 28 ms, 内存消耗: 16.1 MB, 提交时间: 2022-07-09 14:02:12

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        prev = None
        curr = head
        while curr is not None:
            next = curr.next
            curr.next = prev
            prev = curr
            curr = next
        
        return prev

golang 解法, 执行用时: 0 ms, 内存消耗: 2.5 MB, 提交时间: 2021-07-21 10:48:10

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
	var prev *ListNode
	curr := head
	for curr != nil {
		next := curr.Next
		curr.Next = prev
		prev = curr
		curr = next
	}
	return prev
}

golang 解法, 执行用时: 0 ms, 内存消耗: 3 MB, 提交时间: 2021-07-20 10:01:11

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
	if head == nil || head.Next == nil {
		return head
	}
	p := reverseList(head.Next)
	head.Next.Next = head
	head.Next = nil
	return p
}

python3 解法, 执行用时: 44 ms, 内存消耗: 18.3 MB, 提交时间: 2020-11-24 10:18:02

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        # 递归解法
        if head is None or head.next is None:
            return head
        
        p = self.reverseList(head.next)
        head.next.next = head
        head.next = None
        return p
        

python3 解法, 执行用时: 44 ms, 内存消耗: 14.2 MB, 提交时间: 2020-11-24 10:08:32

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        # 迭代解法
        p, rev = head, None
        while p:
            rev, rev.next, p = p, rev, p.next
        return rev