问题描述：
Given an array of integers nums, write a method that returns the "pivot" index of this array.
We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the index.
If no such index exists, we should return -1. If there are multiple pivot indexes, you should return the left-most pivot index.
示例:
Input:
nums = [1, 7, 3, 6, 5, 6]
Output: 3
Explanation:
The sum of the numbers to the left of index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3.
Also, 3 is the first index where this occurs.
Input:
nums = [1, 2, 3]
Output: -1
Explanation:
There is no index that satisfies the conditions in the problem statement.
问题分析：
本题主要看第i的元素之前和之后的和是否相等，是的话返回i，若是不存在，则返回-1.
过程详见代码：
class Solution {
public:
int pivotIndex(vector<int>& nums) {
int sum = accumulate(nums.begin(), nums.end(), 0);
int ts = 0;
for (int i = 0; i < nums.size(); i++)
{
if (sum - nums[i] == 2*ts) return i;
ts += nums[i];
}
return -1;
}
};
