問題很簡單，就是反轉一個 Linked List．

問題

https://leetcode.com/problems/reverse-linked-list/

Given the head of a singly linked list, reverse the list, and return the reversed list.

Example 1:
Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]

Example 2:
Input: head = [1,2]
Output: [2,1]

Example 3:
Input: head = []
Output: []

Constraints:

• The number of nodes in the list is the range [0, 5000].
• -5000 <= Node.val <= 5000

Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?

複習

在 Go 語言，每個變數都要有對應的資料格式，通常我們會使用類型推斷（Type Inference）來宣告變數並給予初始值，譬如：message := "Hello World"．
但如果要宣告一個變數為 nil 值時，則無法使用Type Inference．
所以這個語法previous := nil是錯誤的．
必須將對應的資料格式一起宣告 var previous *ListNode = nil．

我的答案

使用三個指標：

• head: 下一個節點（也可以令外宣告一個 next 指標，但為了節省記憶體，直接用現成的 head 指標．
• current: 目前的節點
• previous: 前一個節點

Python

# 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: Optional[ListNode]) -> Optional[ListNode]:
        previous = None
        current = head
        while current:
            head = head.next
            current.next = previous
            previous = current
            current = head
        return previous

Go

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