二分查找

def binary_search(alist,num):
    n = len(alist)
    
    if len(alist) == 0:
        return False
    mid = n // 2
    
    if alist[mid] == num:
        return True
    elif alist[mid] < num:
        return binary_search(alist[mid+1:],num)
    else: 
        return binary_search(alist[0:mid],num)

if __name__ == "__main__":
    li = list(range(1,40,4))
    
    
    print(binary_search(li,23))


# def binary_search(alist, item):
#     if len(alist) == 0:
#         return False
#     else:
#         midpoint = len(alist)//2
#         if alist[midpoint] == item:
#           return True
#         else:
#           if item < alist[midpoint]:
#             return binary_search(alist[:midpoint],item)
#           else:
#             return binary_search(alist[midpoint+1:],item)

def binary_search(alist,num,left=0,right=0):
    n = len(alist)

    if len(alist) == 0:
        return [left,right]
    mid = n // 2

    if alist[mid] == num:
        return True
    elif alist[mid] < num:
        left = alist[mid]
        return binary_search(alist[mid+1:],num,left,right)
    else:
        right = alist[mid]
        return binary_search(alist[0:mid],num,left,right)

1.最直接的就是 res = min（li），但这多半被扣分，
2.直接遍历，冒泡得到最小的，O（n），效果一般般，
3.前面他问这么多二分查找，只用二分查找得到转折的区间，那个小的数就是最小数了
li[mid] 和 li[-1] 进行比较，37 > 17,说明最小数在 li[mid+1:]
反之说明在 li[:mid+1] 这里的 mid+1 要取到，不然可能遗漏了一点

def minnums(ali):
    if len(ali) == 0:
        return
    elif len(ali) == 1:
        return ali
    elif len(ali) == 2:
        if ali[0] < ali[1]:
            return ali[0]
        else:
            return ali[1]
    else:
        mid = len(ali) // 2
        if ali[mid] > ali[-1]:
            return minnums(ali[mid+1:])
        else:
            return minnums(ali[:mid+1])