1.利用二分法实现O(log n)的查找，但是要主要当index1 = index2 = midindex时，只能按顺序查找

```class Solution:
def minNumberInRotateArray(self, rotateArray):
# write code here
if not rotateArray:return 0
lenth = len(rotateArray)
index1 = 0
index2 = lenth - 1
midindex = 0
while rotateArray[index1] >= rotateArray[index2]:
if index2 - index1 == 1:
midindex = index2
return rotateArray[midindex]
midindex = (index1 + index2) // 2
if rotateArray[index1] == rotateArray[index2] == rotateArray[midindex]:
return self.MinOrder(rotateArray, index1, index2)
if rotateArray[index1] <= rotateArray[midindex]:
index1 = midindex
elif rotateArray[index2] >= rotateArray[midindex]:
index2 = midindex
return rotateArray[midindex]
def MinOrder(self, rotateArray, index1, index2):
result = rotateArray[index1]
for i in range(index1,index2,1):
if rotateArray[i] < result:
result = rotateArray[i]
return result```
1. mid指针值只和right指针指向的值比较大小
• 当 numbers[mid] < numbers[right]​ 时，说明最小值在 ​[left, mid]​ 区间中，则令 right = mid，用于下一轮计算
• 当 numbers[mid] > numbers[right]​ 时，说明最小值在 [mid, right]​ 区间中，则令 left = mid + 1，用于下一轮计算
• 当 numbers[mid] == numbers[right]​ 时，无法判断最小值在哪个区间之中，此时让 right--，缩小区间范围，在下一轮进行判断
```class Solution:
def minArray(self, numbers: List[int]) -> int:
if not numbers:
return None
s = 0
e = len(numbers)-1
while s<e:
mid = (s+e)//2
if numbers[mid] > numbers[e]:
s = mid+1
elif numbers[mid] < numbers[e]:
e = mid
else:
e = e-1
return numbers[s]```