Python return None instead of value

Question

I am trying to implement a very easy Binary search question here: return the index of an array that is plus/minus one from the given target.

However, when I execute this code, my result is always None. Python does not seem to capture my return value.

Here's my code:

class Solution:
    def closestNumber(self, A, target):
        if len(A) == None:
            return -1

        if target - 1 > A[-1] or target + 1 < A[0]:
            return -1

        result = self.BinarySearch(0,len(A), A, target)
        return result
    
    def BinarySearch(self,startIndex, endIndex, A, target):
        if startIndex > endIndex:
            return -1
            
        mid = (startIndex + endIndex)//2
        print(mid)
        if target == A[mid]:
            return mid
        
        if target + 1 == A[mid] or target - 1 == A[mid]:
            return mid
            
        if target + 1 > A[mid]:
            self.BinarySearch(mid, endIndex, A, target)

        if target - 1 < A[mid]:
            self.BinarySearch(startIndex, mid, A, target)
            

    print(Solution().closestNumber([1,4,6,10,20],21))

You're not returning the result of the recursive call. Also your indentation is way off. — deceze, Jan 16 '18 at 10:25

score 1 · Accepted Answer · answered Jan 16 '18 at 10:26

1

You lack 2 return statements in the recursive call branches of your code:

if target + 1 > A[mid]:
    return self.BinarySearch(mid, endIndex, A, target)

if target - 1 < A[mid]:
    return self.BinarySearch(startIndex, mid, A, target)

answered Jan 16 '18 at 10:26

Netwave

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Apparently I was not returning anything from these two branches which was both returning None. Thanks Netwave! – Amanda Zhu Jan 16 '18 at 15:13
1

@AmandaZhu, every python function returns None by default if they don’t return anything explicitly – Netwave Jan 16 '18 at 16:58

Python return None instead of value

1 Answers1