题目：
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.
Example 1:
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.
Example 2:
Input: nums = [1, 2, 3] Output: -1 Explanation: There is no index that satisfies the conditions in the problem statement.
Note:
- The length of
numswill be in the range[0, 10000]. - Each element
nums[i]will be an integer in the range[-1000, 1000].
解题：
如果能找到一个位置使左右两边数字之和相等，返回该位置的索引， 否则返回-1， 计算数组和total， 然后累加数组元素计算sum，计算遍历到的位置，total减去当前的位置数字看下是否未sum的两倍。
实现：
int pivotIndex(int* nums, int numsSize) {
int total = 0;
for (int i=0; i<numsSize; i++){
total += nums[i];
}
int sum=0;
for (int i=0; i<numsSize; i++){
if ((sum + sum) == total - nums[i]) {
return i;
}
sum += nums[i];
}
return -1;
}
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