【Leetcode】python – [189] Rotate Array 個人解法筆記 (內含範例程式碼)

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題目出處

189. Rotate Array

難度

Medium

題目分類

array, math, two pointers

個人範例程式碼

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        len_nums = must_visited = len(nums)
        start_idx = 0
        while(must_visited != 0):
            i = start_idx
            prev = nums[start_idx]
            while(True):
                next_i = (i+k)%len_nums
                save = nums[next_i] # save next
                nums[next_i] = prev # use prev to change next
                prev = save # next give to keep
                i = next_i # move to next index
                must_visited -= 1 # prevent for len_nums can divide k
                if i == start_idx: # return to start 
                    break
            start_idx += 1 # prevent for len_nums can divide k

Time Complexity

O(n)

Space Complexity

O(1) # 這題有特別要求希望是這個空間

算法說明

題目要求空間 O(1),通常就表示這個題目「高機率」可以用 swap 的方式解出來 (想辦法只用交換的)

題目提到「in-place」,表示這題目可以用 swap 的方法解的「可能性很高」,不過這個提示沒有上方機率那麼高,有時候我們也是可以另外 assign 空間再去進行「in-place」的動作。

方法一 – circular array

直覺上,我們可以將此 array 當作一個環,不斷的交換一個循環,直到回到起點結束交換。

為了方便起見,以下的例子我們故意將 index 與裡面的元素值取相同。

我們可以觀察,k=2 時,我們下一個移動的目標 next_i = (i+k)%(array長度) <- 取餘數

我們再次觀察,k=3 時,我們下一個移動的目標 next_i = (i+k)%(array長度) <- 取餘數

注意!! 此方法有個陷阱!!

當「矩陣的長度可以被 k 整除時,會有快速循環的現象」,
例如「len(nums) = 4, k = 2」

[0,1,2,3]
如果只照上述循環後,我們就只會變更 [2,1,0,3] <- 注意只有 0, 2 交換。

為了避免此問題,我們需要另外設計一個 visited 的機制,確保每一個位置都有換過!!

範例程式碼

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        len_nums = must_visited = len(nums)
        start_idx = 0
        while(must_visited != 0):
            i = start_idx
            prev = nums[start_idx]
            while(True):
                next_i = (i+k)%len_nums
                save = nums[next_i] # save next
                nums[next_i] = prev # use prev to change next
                prev = save # next give to keep
                i = next_i # move to next index
                must_visited -= 1 # prevent for len_nums can divide k
                if i == start_idx: # return to start 
                    break
            start_idx += 1 # prevent for len_nums can divide k

corner case 特殊情況處理

處理當 len(nums) = 0 或 k 的情況,基本上都不用做事情,可以節省時間。

上述範例沒有做,但程式碼已經可以處理這個情況。

Boundary conditions/ Edge conditions 邊際情況處理

  • 結束位置在,當 index 返回出發點時。
  • 但要讓整個交換完成,需要注意是否有整除問題,所以真正結束是在 visited 剛好等於 array 長度時,
    期間我們需要去變換起始的 index 位置。

方法二 – recursive

這個方法我也是參考 discussion 看到的,非常酷的做法!
他運用的概念是「矩陣反轉兩次後,會回到原位置」。

因此如果在第二步驟,我們依照「目標 k 位置」進行「分別反轉」,
就可以得到答案。

舉例:k = 3,以下為了方便起見,直接讓 index 值等於元素值

[0,1,2,3,4,5,6] <- 原始位置
[6,5,4,3,2,1,0] <- 第一次 全反轉
[4,5,6,0,1,2,3] <- 拆兩次反轉 0~k-1, k~len(nums)-1

範例程式碼

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        if len(nums) <= 1:
            return
        else:
            k = k % len(nums)
            self.reverse(nums, 0, len(nums)-1)
            self.reverse(nums, 0, k-1)
            self.reverse(nums, k, len(nums)-1)
            return

    def reverse(self, nums, start, end):
        while start < end:
            nums[start], nums[end] = nums[end], nums[start] 
            start += 1
            end -= 1

corner case 特殊情況處理

  • 處理當 len(nums) = 0 或 k 的情況,基本上都不用做事情,可以節省時間。

上述範例沒有做,但程式碼已經可以處理這個情況。
當 0 or k 時,就是 list 進行全部反轉,再反轉一次回來的意思。

  • 處理當 len(nums) = 1 的情況,會超出範圍

新增以下片段

if len(nums) <= 1:
    return
  • 處理當 k > len(nums) 的情況,會超出範圍

這個就是考這題要細心的地方,因為題目沒說 k 必小於 len(nums),
通常會沒考慮到這點,也是要看到測資才會想到…

新增以下片段

k = k % len(nums)

Boundary conditions/ Edge conditions 邊際情況處理

k=3 時,

  • 前半段位置從 0 到 2 = (k-1)
  • 後半段則是 k 開始至 len(nums)-1

Reference

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