二分法有幾種變形：

585  山脈序列中的最大值

first position of target 457 Classical Binary Search 458 Last Position of Target

他們的區別在於：

mountainSequence
二分也不能太死板呀，邊界可以根據具體問題具體分析。
一定要給left = m1 + 1，要不然會超時
最後返回這個
            else:
                left = m1
                right = m2
            
        return max(nums[left], nums[right])


first position of target
只有兩個if else，沒有第三個if


457 Classical Binary Search
任意找一個
            else:
                return mid


458 Last Position of Target
            else:
                start = mid 
 

用圖片展示區別

四種方法程式：

 二分查詢 · first position of target  

class Solution:
    # @param nums: The integer array
    # @param target: Target number to find
    # @return the first position of target in nums, position start from 0 
    def binarySearch(self, nums, target):
        # write your code here
        left, right = 0, len(nums)
        while left + 1 < right :
            mid = (left + right) // 2
            if nums[mid] < target :
                left = mid
            else :
                right = mid   #第一次，最後一次，隨意三種二分查詢都是在這裡變化了


        if nums[left] == target :
            return left
        elif nums[right] == target :
            return right
        return -1;
 

457 Classical Binary Search  

#隨便找一個就行
class Solution:
    """
    @param nums: An integer array sorted in ascending order
    @param target: An integer
    @return: An integer
    """
    def findPosition(self, nums, target):
        # write your code here
        if not nums or target is None:
            return -1
            
        start, end = 0, len(nums) - 1
        while start + 1 < end:
            mid = (end + start)//2
            if target < nums[mid]:
                end = mid
            elif target > nums[mid]:
                start = mid
            else:
                return mid
        
        if target == nums[start]:
            return start
        if target == nums[end]:
            return end 
        return -1
    



                    
my_solution = Solution()
A = [1, 2, 2, 4, 5, 5]
target = 2
result = my_solution.findPosition(A, target)
print('result is',result)     
 

458 Last Position of Target  

class Solution:
    # @param {int[]} A an integer array sorted in ascending order
    # @param {int} target an integer
    # @return {int} an integer
    def lastPosition(self, A, target):
        if not A or target is None:
            return -1

        start = 0
        end = len(A) - 1

        while start + 1 < end:  #start為啥加1？ 為啥不是小於等於； 使用 start < end 無論如何都會出現死迴圈
            mid = start + (end - start) // 2  #避免記憶體溢位

            if A[mid] < target:
                start = mid
            elif A[mid] > target:
                end = mid
            else:
                start = mid  #A[mid] == target:   #return mid 是隨意返回一箇中間位置，這裡要求最後一個位置，所以放棄這種寫法

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

my_solution = Solution()
A = [1, 2, 2, 4, 5, 5]
target = 2
result = my_solution.lastPosition(A, target)
print('result is',result)

585  山脈序列中的最大值  

class Solution:
    # @param {int[]} nums a mountain sequence which increase firstly and then decrease
    # @return {int} then mountain top
    def mountainSequence(self, nums):
        # Write your code here
        left, right = 0, len(nums) - 1
        while left + 1 < right:
            m1 = left + (right - left) // 2  # m1 is middle
            #m1 = right - (right - left) // 2
            m2 = right - (right - m1) // 2      #m2 = (right + m1)//2 是右側和陣列中點的中點
            #m2 = (right + m1)//2  #這種寫法會報錯
            if nums[m1] < nums[m2]:
                left = m1 + 1
            elif nums[m1] > nums[m2]:
                right = m2 - 1
            else:
                left = m1
                right = m2
            
        return max(nums[left], nums[right])

my_solution = Solution()
A = [1, 2, 4, 8, 6, 3]
result = my_solution.mountainSequence(A)
print('result is',result)

認識你是我們的緣分，同學，等等，記得關注我。

微信掃一掃 關注該公眾號