724. Find Pivot Index

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 nums will be in the range [0, 10000].
Each element nums[i] will be an integer in the range [-1000, 1000].

这题给了限制条件，所以很简单，所有数的和不会超过32 bit int 的上下限（21亿左右）。
所以这题直接对所有数求和，然后从左往右重新遍历，找到第一个左右和相等的index。

int pivotIndex(vector<int>& nums) {    int sum = 0;    for (int i = 0; i < nums.size(); i++) {        sum += nums[i];    }    int tempsum = 0;    for (int i = 0; i < nums.size(); i++) {        sum = sum - nums[i];        if (tempsum == sum) {            return i;        }        tempsum += nums[i];    }    return -1;}