binary search implementation with python

问题

I think I did everything correctly, but the base case return None, instead of False if the value does not exists. I cannot understand why.

def binary_search(lst, value):
    if len(lst) == 1:
        return lst[0] == value

    mid = len(lst)/2
    if lst[mid] < value:
        binary_search(lst[:mid], value)
    elif lst[mid] > value:
        binary_search(lst[mid+1:], value)
    else:
        return True

print binary_search([1,2,4,5], 15)

回答1:

You need to return the result of the recursive method invocation:

def binary_search(lst, value):
    #base case here
    if len(lst) == 1:
        return lst[0] == value

    mid = len(lst)/2
    if lst[mid] < value:
        return binary_search(lst[:mid], value)
    elif lst[mid] > value:
        return binary_search(lst[mid+1:], value)
    else:
        return True

And I think your if and elif condition are reversed. That should be:

if lst[mid] > value:    # Should be `>` instead of `<`
    # If value at `mid` is greater than `value`, 
    # then you should search before `mid`.
    return binary_search(lst[:mid], value)
elif lst[mid] < value:  
    return binary_search(lst[mid+1:], value)

回答2:

Because if return nothing!

if lst[mid] < value:
    binary_search(lst[:mid], value)
    # hidden return None
elif lst[mid] > value:
    binary_search(lst[mid+1:], value)
    # hidden return None
else:
    return True

回答3:

You need to return from if and elif too.

def binary_search(lst, value):
    #base case here
    if len(lst) == 1:
        return lst[0] == value

    mid = len(lst) / 2
    if lst[mid] < value:
        return binary_search(lst[:mid], value)
    elif lst[mid] > value:
        return binary_search(lst[mid+1:], value)
    else:
        return True

>>> print binary_search([1,2,4,5], 15)
False

回答4:

def binary_search(lists,x):
    lists.sort()
    mid = (len(lists) - 1)//2
    if len(lists)>=1:
        if x == lists[mid]:
            return True

        elif x < lists[mid]:
            lists = lists[0:mid]
            return binary_search(lists,x)

        else:
            lists = lists[mid+1:]
            return binary_search(lists,x)
    else:
        return False
a = list(map(int,input('enter list :').strip().split()))
x = int(input('enter number for binary search : '))
(binary_search(a,x))

回答5:

def rBinarySearch(list,element):
    if len(list) == 1:
        return element == list[0]
    mid = len(list)/2
    if list[mid] > element:
        return rBinarySearch( list[ : mid] , element )
    if list[mid] < element:
        return rBinarySearch( list[mid : ] , element)
    return True

来源：https://stackoverflow.com/questions/18010660/binary-search-implementation-with-python

标签

python

algorithm

python-2.x