27 函数的递归

Posted 2021-01-08 bydzxzy

# 二分查找算法 必须处理有序的列表
# l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88]

def find(l,aim):
    mid_index = len(l) // 2
    if l[mid_index] < aim:
        new_l = l[mid_index+1 :]
        find(new_l,aim)
    elif l[mid_index] > aim:
        new_l = l[:mid_index]
        find(new_l, aim)
    else:
        print(‘找到了‘,mid_index,l[mid_index])

find(l,66)

但是，存在好多问题，所以

l = [2,3,5,10,15,16,18,22,26,30,32,35,41,42,43,55,56,66,67,69,72,76,82,83,88]
def find(l,aim,start = 0,end = None):
    end = len(l) if end is None else end
    mid_index = (end - start)//2 + start
    if start <= end:
        if l[mid_index] < aim:
            return find(l,aim,start =mid_index+1,end=end)
        elif l[mid_index] > aim:
            return find(l, aim, start=start, end=mid_index-1)
        else:
            return mid_index
    else:
        return ‘找不到这个值‘

ret= find(l,44)
print(ret)