題目出處

33. Search in Rotated Sorted Array

難度

Medium

題目分類

Array, BinarySearch

難度

medium

個人範例程式碼 – 2022/6/7 三刷

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        start = 0
        end = len(nums) - 1

        while start + 1 < end:
            mid = (start + end) // 2
            # print(start, mid, end)
            # print(nums[start], nums[mid], nums[end])

            if nums[mid] == target:
                return mid

            if nums[start] < nums[mid]: # front: simple go up
                if nums[start] <= target < nums[mid]:
                    end = mid   
                else:
                    start = mid
            else: # back: simple go up
                if nums[mid] < target <= nums[end]:
                    start = mid
                else:
                    end = mid

        else:
            if nums[start] == target:
                return start
            elif nums[end] == target:
                return end
            else:
                return -1

最近在練習程式碼本身就可以自解釋的 Coding style，可以嘗試直接閱讀程式碼理解

算法說明

分成 2 種 case 討論，2 種 case 底下又有兩種 case，

先看 mid 的落點，
分成 「start < mid」 或 「mid < end」

再來各自看 target 的落點，(看有沒有落在單純上升的範圍)。

• 最後我們可以整理成：

• 「start < mid」

  • 「start < target < mid」：純上升範圍
  • 「else」：亂的範圍 (依然是 rotated sorted array)
• 「mid < end」
  • 「mid < target < end」：純上升範圍
  • 「else」：亂的範圍 (依然是 rotated sorted array)

input handling

處理沒有輸入的時候，return -1

Boundary conditions

binary search 的結束條件

個人範例程式碼 – 2022/4/5 二刷

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        if not nums:
            return -1

        start, end = 0, len(nums)-1
        while(start + 1 < end):
            mid = (start + end) // 2
            if(nums[mid] >= nums[end]):
                if(nums[start] <= target <= nums[mid]):
                    end = mid
                else:
                    start = mid
            else: # (nums[mid] < nums[end]):
                if(nums[mid] <= target <= nums[end]):
                    start = mid
                else:
                    end = mid

        else:
            if(nums[start] == target):
                return start
            elif(nums[end] == target):
                return end
            else:
                return -1

最近在練習程式碼本身就可以自解釋的 Coding style，可以嘗試直接閱讀程式碼理解

算法說明

這題可以說是 binary search 的最 #重要問題，
主要因為要寫得出這題，基本上要對 binary search 的各種細節幾乎都已經非常清楚才寫得出來。

注意幾個討論的重點，我們可以把整個題目拆成 2*2 個 case，

• 比 start 還大的情況 (等同於 >= end 的情況)
• 比 end 還小的情況

然後再針對這兩個情況，再拆成兩種 mid 的情況：

• 比 start 還大的情況 (等同於 >= end 的情況)
  • start < target < mid 的情況
  • else
• 比 end 還小的情況
  • else
  • mid < target < end 的情況

我們要注意移動的時候要移動 start 還是 end，可以從圖上觀察而出。

此外，邊界條件的處理也是此題相當重要的關鍵。

「>= end」

這個觀念我們在 【Leetcode】python – [153] Find Minimum in Rotated Sorted Array 個人解法筆記
已經有詳細討論過，如果不清楚可以去看看

「邊界有沒有等於」?

這問題也非常重要，因為這會影響我們要移動 start 或是 end，
而我們需要讓邊界「等於」，因為如果 mid 正好是解答，

我們需要讓「移動側 = mid = target」，因此邊界條件中會有「包含等於的條件」

input handling

處理沒有輸入的時候，return -1

Boundary conditions

binary search 的結束條件

個人解法筆記 (解法重點) – 2021/7/7 一刷

示意圖

注意事項

注意 corner case 的處理
範例： [5, 1, 3]

需要注意判斷非 ascending 時，是否該值會出現在對應的區間。

個人範例程式碼

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        l_idx , r_idx = 0, len(nums)-1

        while l_idx <= r_idx:
            mid_idx = (l_idx + r_idx)//2

            if nums[mid_idx] == target:
                return mid_idx

            # search left
            if nums[l_idx] <= nums[mid_idx]:
                if nums[l_idx] <= target and target < nums[mid_idx]: # ascending side
                    r_idx = mid_idx - 1
                else:
                    l_idx = mid_idx + 1

            # search right
            else: 
                if nums[mid_idx] < target and target <= nums[r_idx]: # ascending side
                    l_idx = mid_idx + 1
                else:
                    r_idx = mid_idx - 1


        return -1

Reference

• 搜索旋转排序数组 · Search in Rotated Sorted Array
• https://leetcode.com/problems/remove-nth-node-from-end-of-list/discuss/8802/3-short-Python-solutions

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