【Python】抛磚引玉-連續有序數組的排序問題

有這麼一個連續有序數組，比如1,2,3,4,5,6,7,8,9。把7,8,9移到最前面，就是7,8,9,1,2,3,4,5,6。讓你找出某個元素在數組中的索引，如果沒有則傳回-1

用python我的解法是這樣的，先把那個分界點找出來，然後切成兩個數組，判斷下該在哪個數組裡進行二分查找，然後進行二分查找：

# Online Python compiler (interpreter) to run Python online.
# Write Python 3 code in this online editor and run it.
def find_sub_arr(_arr):
    sub_arr1 = []
    sub_arr2 = []
    i = 0
    while i < len(_arr) - 1:
        if _arr[i+1] != _arr[i] + 1:
            sub_arr1 = _arr[:i+1]
            sub_arr2 = _arr[i+1:]
        i += 1
    return sub_arr1, sub_arr2

def binary_search(_arr, l, r, _num):
    if r >= l:
        mid = int(l + (r - l)/2)
        if _arr[mid] == _num:
            return mid
        elif _arr[mid] > _num: 
            return binary_search(_arr, l, mid-1, _num)
        else:
            return binary_search(_arr, mid+1, r, _num)
    else:
        return -1
    
def index_of(_arr, _num):
    sub1, sub2 = find_sub_arr(arr)
    if _num >= sub1[0]:
        print(f"find {_num} in {sub1}")
        return binary_search(sub1, 0, len(sub1)-1, _num)
    elif _num >= sub2[0]:
        print(f"find {_num} in {sub2}")
        return len(sub1) + binary_search(sub2, 0, len(sub2)-1, _num)
    else:
        return -1
    
arr = [7,8,9,1,2,3,4,5,6]
print(f"arr is {arr}")
for a in arr:
    ind = index_of(arr, a)
    print(f"index {a} is {ind}")
    print("——————————————————")