二分法查找

二分查找--非递归算法
# lst = [22,33,44,55,66,77,88,99,101,238,345,456,567,678,789]
# n=238
# left = 0
# right = len(lst)-1
# count = 1
# while left<=right:
#     middle = (left+right)//2
#     if n <lst[middle]:
#         right = middle-1
#     elif n > lst[middle]:
#         left = middle+1
#     else:
#         print(count)
#         print(middle)
#         break
#     count = count + 1
# else:
#     print("不存在")

普通递归二分法查找
def binary_search(n,left,right):
    if left<=right:
        middle = (left+right)//2
        if n < lst[middle]:
            right = middle -1
        elif n > lst[middle]:
            left= middle+1
        else:
            return middle
        return binary_search(n,left,right)
    else:
        return -1
print(binary_search(567,0,len(lst)-1))

#递归 二分法,很难计算位置
# def binary_search(ls,target):
#     left = 0
#     right = len(ls) - 1
#     if left > right:
#         print("不在这里")
#     middle = (left+right)//2
#     if target< ls[middle]:
#         return binary_search(ls[:middle],target)
#     elif target > ls[middle]:
#         return binary_search(ls[middle+1:],target)
#     else:
#         print("在这里")

#递归二分法2
def search(l,aim,start=0,end=None):
    if end is None:end = len(l)-1
    if  start<=end:
        mid = start + (end-start)//2
        mid_value = l[mid]
        if aim > mid_value:
            start = mid+1
            ret = search(l,aim,start,end)
            return ret
        elif aim < mid_value:
            end = mid-1
            ret = search(l,aim,start,end)
            return ret
        else:
            return mid
    else:
        pass