问题
I have a list with numbers from 0-9:
mylist = list(range(10))
I am getting an error with the division command to get mid:
def binary_search(mylist, element, low, high):
low=0
high= len(mylist)
mid=low + (high- mymin)/2
if mid==len(mylist):
return False
elif mylist[mid]==element:
return mid
elif high==low:
return False
elif mylist[mid]<element:
return binary_search(mylist, element, mymin, mid-1)
elif mylist[mid]<element:
return binary_search(mylist, element, mid+1, mymax)
else:
return mid
and if I wanted to return True how would I write that on top of return binary_search(mylist, element, mymin, mid-1)?
回答1:
Recursive solution:
def binary_search_recursive(arr, elem, start=0, end=None):
if end is None:
end = len(arr) - 1
if start > end:
return False
mid = (start + end) // 2
if elem == arr[mid]:
return mid
if elem < arr[mid]:
return binary_search_recursive(arr, elem, start, mid-1)
# elem > arr[mid]
return binary_search_recursive(arr, elem, mid+1, end)
Iterative solution:
def binary_search_iterative(arr, elem):
start, end = 0, (len(arr) - 1)
while start <= end:
mid = (start + end) // 2
if elem == arr[mid]:
return mid
if elem < arr[mid]:
end = mid - 1
else: # elem > arr[mid]
start = mid + 1
return False
回答2:
The first solution looks wrong because it doesn't index the list.
This problem tripped me up too the first time I wrote a solution so be sure to test your algorithm well.
Here's what I ended up with:
def binary_search(value, items, low=0, high=None):
"""
Binary search function.
Assumes 'items' is a sorted list.
The search range is [low, high)
"""
high = len(items) if high is None else high
pos = low + (high - low) / len(items)
if pos == len(items):
return False
elif items[pos] == value:
return pos
elif high == low:
return False
elif items[pos] < value:
return binary_search(value, items, pos + 1, high)
else:
assert items[pos] > value
return binary_search(value, items, low, pos)
And when I test it, the answers look correct:
In [9]: for val in range(7):
...: print val, binary_search(val, [1, 2, 3, 5])
...:
0 False
1 0
2 1
3 2
4 False
5 3
6 False
Btw, Python has a library module for just this kind of thing named bisect.
回答3:
You can make use of list slicing too.
def binary_search_recursive(list1, element):
if len(list1) == 0:
return False
else:
mid = len(list1)//2
if (element == list1[mid]):
return True
else:
if element > list1[mid]:
return binary_search_recursive(list1[mid+1:],element)
else:
return binary_search_recursive(list1[:mid],element)
However, note that list slicing introduces additional complexity.
回答4:
There are a lot of solutions here already. Below is one more solution without slicing and that just requires element and list as arguments:
def binary_search(item, arr):
def _binary_search(item, first, last, arr):
if last < first:
return False
if last == first:
return arr[last] == item
mid = (first + last) // 2
if arr[mid] > item:
last = mid
return _binary_search(item, first, last, arr)
elif arr[mid] < item:
first = mid + 1
return _binary_search(item, first, last, arr)
else:
return arr[mid] == item
return _binary_search(item, 0, len(arr) - 1, arr)
print(binary_search(-1, [0, 1, 2, 3, 4, 5]))
回答5:
please correct me, if the code presents any bugs
def binary_recursive(array, val):
if val < array[0] or val > array[-1]:
return False
mid = len(array) // 2
left = array[:mid]
right = array[mid:]
if val == array[mid]:
return True
elif array[mid] > val:
return binary_recursive(left, val)
elif array[mid] < val:
return binary_recursive(right, val)
else:
return False
回答6:
Your first one won't even get started, because list(mid) will immediately raise a TypeError: 'list' object is not callable.
If you fix that (by using list[mid]), your next problem is that you ignore the min and max arguments you receive, and instead set them to 0 and len(list)-1 each time. So, you will infinitely recurse (until you eventually get a RecursionError for hitting the stack limit).
Think about it: the first call to binarySearch(l, 5, 0, 10) recursively calls binarySearch(l, 5, 0, 4). But that call ignores that 4 and acts as if you'd passed 10, so it recursively calls binarySearch(l, 5, 0, 4). Which calls binarySearch(l, 5, 0, 4). And so on.
If you fix that (by removing those two lines), you've got another problem. When you give it the number 10, binarySearch(l, 10, 0, 10) will call binarySearch(l, 10, 6, 10), which will callbinarySearch(l, 10, 8, 10), then binarySearch(l, 10, 9, 10), thenbinarySearch(l, 10, 10, 10). And that last one will check list[10] > element. But list[10] is going to raise an IndexError, because there aren't 11 elements in the list.
And once you fix that off-by-one error, there are no problems left. The problem you asked about cannot possibly ever occur. Here's an example run:
>>> a = range(10)
>>> for i in -3, -1, 0, 1, 4, 5, 9, 10, 11:
... print i, binarySearch(a, i, 0, 10)
-3 False
-1 False
0 0
1 1
4 4
5 5
9 9
10 False
11 False
Your second version isn't recursive. bSearch never calls bSearch, and that's the whole definition of recursion.
There's nothing wrong with writing an iterative algorithm instead of a recursive algorithm (unless you're doing a homework problem and recursion is the whole point), but your function isn't iterative either—there are no loops anywhere.
(This version also ignores the start and end arguments, but that's not as much of a problem in this case, because, again, you're not doing any recursion.)
Anyway, the only return False in the function is in that first if len(list) == 0. So, for any non-empty list, it's either going to return True, or a number. With your input, it will return 4 for anything less than the value at the midpoint of the list (5), and True for anything else.
回答7:
Your problem here is that you're redeclaring min and max in each loop, so although it should be recursive, passing in a new min or max each time, this isn't in fact happening.
You can solve this by using defaults in the arguments:
def binary_search(list, element, min=0, max=None):
max = max or len(list)-1
if max<min:
return False
else:
mid= min + (max-min)/2
if mid>element:
return binary_search(list, element, min, mid-1)
elif mid<element:
return binary_search(list, element, mid+1, max)
else:
return mid
If you're not familiar with the syntax on line 2, max = max or len(list)-1 max will be set to len(list)-1 only if max is not passed in to the method.
So you can call the method simply using:
binary_search(range(10), 7)
# Returns 7
binary_search(range(10), 11)
# Returns False
回答8:
Just another answer to the same question:
def binary_search(array, element, mini=0, maxi=None):
"""recursive binary search."""
maxi = len(array) - 1 if maxi is None else maxi
if mini == maxi:
return array[mini] == element
else:
median = (maxi + mini)/2
if array[median] == element:
return True
elif array[median] > element:
return binary_search(array, element, mini, median)
else:
return binary_search(array, element, median+1, maxi)
print binary_search([1,2,3],2)
回答9:
I made this one. Correct me if there's any bug.
import math
def insert_rec(A,v,fi,li):
mi = int(math.floor((li+fi)/2))
if A[mi] == v:
print("Yes found at: ",mi)
return
if fi==li or fi>li:
print("Not found")
return
if A[mi] < v:
fi = mi+1
insert_rec(A,v,fi,li)
if A[mi] > v:
li = mi-1
insert_rec(A,v,fi,li)
回答10:
def bs(list,num): #presume that the list is a sorted list
#base case
mid=int(len(list)/2) # to divide the list into two parts
if num==list[mid]:
return True
if len(list)==1:
return False
#recursion
elif num<list[mid]: #if the num is less than mid
return bs(list[0:mid],num) #then omit all the nums after the mid
elif num>list[mid]: #if the num is greater than mid
return bs(list[mid:],num) # then omit all the nums before the mid
#return False
list=[1,2,3,4,5,6,7,8,9,10]
print(bs(list,20))
<<< False
print(bs(list,4))
<<< True
回答11:
You can implement binary search in python in the following way.
def binary_search_recursive(list_of_numbers, number, start=0, end=None):
# The end of our search is initialized to None. First we set the end to the length of the sequence.
if end is None:
end = len(list_of_numbers) - 1
if start > end:
# This will happen if the list is empty of the number is not found in the list.
raise ValueError('Number not in list')
mid = (start + end) // 2 # This is the mid value of our binary search.
if number == list_of_numbers[mid]:
# We have found the number in our list. Let's return the index.
return mid
if number < list_of_numbers[mid]:
# Number lies in the lower half. So we call the function again changing the end value to 'mid - 1' Here we are entering the recursive mode.
return binary_search_recursive(list_of_numbers, number, start, mid - 1)
# number > list_of_numbers[mid]
# Number lies in the upper half. So we call the function again changing the start value to 'mid + 1' Here we are entering the recursive mode.
return binary_search_recursive(list_of_numbers, number, mid + 1, end)
We should check our code with good unittest to find loop holes in our code.
Hope this helps you.
回答12:
Here's an elegant recursive solution if you're:
1) Trying to return the INDEX of the target in the original list being passed in, before it is halved. Getting the target is the easy part.
2) Only want to have to pass in 2 arguments: (list, target) instead of 4 arguments including the upper/lower (right/left) bounds of each array being passed in recursively.
3) Don't want out of bounds, maximum recursion depth, or target not found errors.
# Base Case: one item (target) in array.
# Recursive Case: cut array by half each recursive call.
def recursive_binary_search(arr, target):
mid = len(arr) // 2
if len(arr) == 1:
return mid if arr[mid] == target else None
elif arr[mid] == target:
return mid
else:
if arr[mid] < target:
callback_response = recursive_binary_search((arr[mid:]), target)
return mid + callback_response if callback_response is not None else None
else:
return recursive_binary_search((arr[:mid]), target)
回答13:
Here is My Recursion Solution of Binary Search
def recBinarySearch(arr,ele):
if len(arr) == 0:
return False
else:
mid = len(arr)/2
if arr[mid] == ele:
return True
else:
if ele < arr[mid]:
return recBinarySearch(arr[:mid], ele)
else:
return recBinarySearch(arr[mid+1], ele)
回答14:
Though it's too late, it might help someone else :-
def bsearch_helper(arr, key, low, high):
if low > high:
return False
mid = (low + high)//2
if arr[mid] == key:
return True
elif arr[mid] < key:
return bsearch_helper(arr, key, mid + 1, high)
else:
return bsearch_helper(arr, key, low, mid - 1)
return False
def bsearch(arr, key):
return bsearch_helper(arr, key, 0, len(arr) - 1)
if __name__ == '__main__':
arr = [8, 3, 9, 2, 6, 5, 1, 7, 4]
print (bsearch(sorted(arr), 5))
来源：https://stackoverflow.com/questions/19989910/recursion-binary-search-in-python