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].

自己的解法

class Solution {
    public int pivotIndex(int[] nums) {
        int right = 0;
        for (int i = 0; i < nums.length; i++) {
            int sumr = 0;
            int suml = 0;
            for (int j = 0; j < i; j++) {
                suml = suml + nums[j];
                right = suml;
            }
            for (int l = i + 1; l < nums.length; l++) {
                sumr = sumr + nums[l];
                
            }
            if (right == sumr) {
                    return i;
                } 
        }
        return -1;
    }
}

假设我们知道S是数字的总和，我们在索引i。如果我们知道索引i左边的数字总和，那么索引右边的另一个总和就是S - nums [i] - leftsum。因此，我们只需要知道leftsum来检查索引是否是恒定时间内的枢轴索引。让我们这样做：当我们迭代候选索引i时，我们将保持leftsum的正确值。

class Solution {
    public int pivotIndex(int[] nums) {
        int sum = 0, leftsum = 0;
        for (int x: nums) sum += x;
        for (int i = 0; i < nums.length; ++i) {
            if (leftsum == sum - leftsum - nums[i]) return i;
            leftsum += nums[i];
        }
        return -1;
    }
}