"在排序数组中查找 Pivot 元素，如何优化此代码？"

Question

Is there a way to optimize this code? I'm trying to find an element in the list where the sorted list is rotated. [8,9,10,11,12,6,7] Here the element I'm trying to find is 6 with an index 5.

class Solution:    
    def search(self, nums: List[int]) -> int:
        lenn = len(nums)-1
        pivot, mid = 0, (0 + lenn)//2
        
        if(nums[0] > nums[lenn]):
            while True:
                if(nums[mid] > nums[mid+1]):
                    pivot = mid+1
                    break
                if(nums[mid] < nums[mid-1]):
                    pivot = mid
                    break
                if(nums[mid] > nums[lenn]):
                    mid += 1
                if(nums[mid] < nums[lenn]):
                    mid -= 1
        return pivot

Answer 1

Your solution is fast for small lists but not for big ones as it perform a linear-time search<\/strong> . To be faster, you can use a binary search<\/strong> . To do that, you need to use a keep a [start;end]<\/code> range of values that contains the "pivot". You just need to take the middle item and find which part (left or right) contains increasing items.

def search(nums):
    first, last = nums[0], nums[-1]

    # Trivial case
    if first <= last:
        return 0

    lenn = len(nums)-1

    # Micro-optimized fast linear search for few items (optional)
    if lenn < 10:
        pivot, mid = 0, lenn//2
        while True:
            middle = nums[mid]
            if middle > nums[mid+1]:
                return mid+1
            elif middle < nums[mid-1]:
                return mid
            elif middle > last:
                mid += 1
            elif middle < last:
                mid -= 1

    # Binary search for many items
    start, end = 0, lenn
    while start+1 < end:
        mid = (start + end) // 2
        midVal = nums[mid]
        if midVal < nums[start]:
            end = mid
        elif midVal > nums[end]:
            start = mid
        else:
            assert False
    return start+1