二分查找非递归版本python实现

def binary_search_2(arr,item):
    n = len(arr)
    mid = n//2

    while mid>=0 and mid<n-1:

        if arr[mid]==item:
            return True
        elif arr[mid]>item:
            mid = mid//2
        else:
            mid = (mid+n)//2
    if arr[-1]==item:
        return True
    return False

#递归转非递归的关键1循环2mid值的转变

if  __name__ == '__main__':
    li = [1,3,6,8,10,12,14,16]
    print(binary_search_2(li,1))
    print(binary_search_2(li,14))
    print(binary_search_2(li,98))

时间O（logn）