Code wars :Equal Sides Of An Array

You are going to be given an array of integers. Your job is to take that array and find an index N where the sum of the integers to the left of N is equal to the sum of the integers to the right of N. If there is no index that would make this happen, return -1.

For example:

Let's say you are given the array {1,2,3,4,3,2,1}:

Your function will return the index 3, because at the 3rd position of the array, the sum of left side of the index ({1,2,3}) and the sum of the right side of the index ({3,2,1}) both equal 6.

Let's look at another one.You are given the array {1,100,50,-51,1,1}:

Your function will return the index 1, because at the 1st position of the array, the sum of left side of the index ({1}) and the sum of the right side of the index ({50,-51,1,1}) both equal 1.

Last one:You are given the array {20,10,-80,10,10,15,35}

At index 0 the left side is {}The right side is {10,-80,10,10,15,35}They both are equal to 0 when added. (Empty arrays are equal to 0 in this problem)Index 0 is the place where the left side and right side are equal.Note: Please remember that in most programming/scripting languages the index of an array starts at 0

我的解法：

def find_even_index(arr):
    leftsum = 0
    rightsum = 0
    for i in range(1,len(arr)) :
        rightsum += arr[i]
    if(rightsum == 0) :
        return 0
    for i in range(1,len(arr)):
        leftsum += arr[i-1]
        rightsum -= arr[i]
        if(leftsum == rightsum):
            return i
    return -1

 

 

大神解法：

def find_even_index(arr):

    for i in range(len(arr)):

        if sum(arr[:i]) == sum(arr[i+1:]):

            return i

    return -1

 

 

个人感觉，自己的算法要少计算几次加法