参考：http://*.com/questions/34584416/nested-loops-with-arrays

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.

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.

Example Tests:

 Test.describe("FindEvenIndex", function() {

   Test.it("Tests", function() {

     Test.assertEquals(findEvenIndex([1,2,3,4,3,2,1]),3, "The array was: [1,2,3,4,3,2,1] \n");

     Test.assertEquals(findEvenIndex([1,100,50,-51,1,1]),1, "The array was: [1,100,50,-51,1,1] \n");

     Test.assertEquals(findEvenIndex([1,2,3,4,5,6]),-1, "The array was: [1,2,3,4,5,6] \n");

     Test.assertEquals(findEvenIndex([20,10,30,10,10,15,35]),3, "The array was: [20,10,30,10,10,15,35] \n");

   });

 });

Solution:

 function findEvenIndex(arr) {

   var add    = (a, b) => a + b;

   var sum    = a => a.reduce(add, 0);

   var isEven = (e, i, a) => sum(a.slice(0, i)) === sum(a.slice(i+1));

   return arr.findIndex(isEven);

 }