Equal Sides Of An Array--codewar刷题总结

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.

Input:
An integer array of length 0 < arr < 1000. The numbers in the array can be any integer positive or negative.

Output:
The lowest index N where the side to the left of N is equal to the side to the right of N. If you do not find an index that fits these rules, then you will return -1.

Note:
If you are given an array with multiple answers, return the lowest correct index.

 看一下我的解决方案，说明一下，做这道题的时候没有思路，网上找了一些资源，思路搞懂了，自己才写出来的。

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

 这个非常好理解，是将每一个i做为边界将数组切片，前后和比较一下。

再看一个。

def find_even_index(arr):
    arr_sum=sum(arr)
    left_sum=0
    for index in range(len(arr)):
        if left_sum*2==arr_sum-arr[index]:
            return index
        left_sum+=arr[index]
    return -1

 这个思路比较不错，也没有调用那么多的内置函数。它的思路是如果遍历到这里，两边相等，那么前面的和乘以2必然等于数组和减arr[i]，随着遍历，前面的和可以慢慢往上递增。

再看一下别人的。

def find_even_index(lst):
    left_sum = 0
    right_sum = sum(lst)
    for i, a in enumerate(lst):
        right_sum -= a
        if left_sum == right_sum:
            return i
        left_sum += a
    return -1

 这个方案先将总和算出来，然后随着遍历，一点一点往下减，另一边一点一点往上加，直到两边一样，值得注意的是arr[i]在这里并不用考虑，虽然sum[arr]将它算进去了，但慢慢往下减的过程中，只要将正确的i减掉，而left_sum往上加的代码在if下面，这时并没有将正确的i加进去，此时它们是相等的，就进入if语句结束了。

非常巧妙的思路。