【Leetcode】python – [206] Reverse Linked List 個人解法筆記 (內含範例程式碼) 內含 Linked List reverse 反轉操作 part 3 (for 新手教學)

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題目出處

206. Reverse Linked List

難度

easy

題目分類

Linked List, Recursion

Linked List reverse 反轉操作 part 3

這邊我們要來研究 Linked list 的 反轉操作,
一樣的,這操作硬要說也不難,但只要思緒不夠清楚,
一定是轉了半天轉不出來XDD 最後絕望放棄XD

我們試著來把概念講的簡單一點吧!

試試看講簡單一點

做反轉動作來說,最簡單我們都需要「三個紀錄點」,
分別代表「過去」、「現在」、「未來」,
每一個記錄點都非常重要,
我們一定要注意「不能在過程中失去任何重要紀錄點,一旦被蓋掉就再也找不回來了!

因此我們基本的策略就有了下圖,1 ~ 3,
我們移動的方式一定是:

  1. 過去 = 現在
  2. 現在 = 未來
  3. 未來 = 更未來

到這邊,「務必務必謹守上述這個原則」,
再來我們再來討論反轉,什麼時候適合做反轉呢?
就是在做上述的「移動 1~3」之前,所以「更改」反轉的線,就是在第 0 步驟,

初始化

這裡有個小重點,不論如何,我們的 Linklist 最終「必須結束於 None」,
因此我們要先宣告個「新結束點」給他
那想當然,可以讓一開始 prev 擔任這個步驟。

而我們真正作為起頭的 head ,可以先給予他 cur (代表現在),
由於此時我們還不確定有沒有 head 存在,因此還不能直接給 head.next (會跳錯)

個人範例程式碼

記得上述所說的,我們需要先宣告 None 作為結尾

# 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]:
        prev = None # must have a None end.
        if head: # init 
            cur = head
            head = head.next
            cur.next = prev # must have a None end.
        else: # special case
            return head

        # loop
        while head:   
            prev = cur
            cur = head  
            head = head.next
            cur.next = prev # reverse

        return cur 

Time Complexity

O(n)

Space Complexity

O(1)

算法說明

遍歷一次 head 是一定要的,問題是我們該怎麼移除,
移除的方法我們上面也提到過了,
只需要把「下個」替換成「下下個」就可以輕鬆解決,
只是要多考慮一些「不存在的邊際情況」,這邊我們下面會提。

corner case 特殊情況處理

一同整理在下方「邊際情況處理」

Boundary conditions/ Edge conditions 邊際情況處理

我們把題目稍微分析一下,我這裡大概把題目分類成於以下三種情況,

  1. [None]
  2. [1]
  3. [1,2,3] <- 正常情況

其中我們來看看 1.2. 我們該怎麼辦。

1. [None]

如果一開始 if head 就沒東西,我們就直接 return head

2. [1]

我們已經把 cur = head,而 while head 會直接判找不到東西,
因此回傳 cur,沒有問題。

Reference

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Howard Weng

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