Formatted question description: https://leetcode.ca/all/724.html

Easy

## Description

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].

## Solution

Maintain the sums in the left subarray and the right subarray. Initially, the candidate pivot index is 0, so the left subarray has length 0 and the right subarray has length nums.length - 1. If the left subarray sum equals the right subarray sum, return the current index as the pivot index. Otherwise, add the candidate pivot index by 1 and update the left subarray sum and the right subarray sum. If at a time the left subarray sum equals the right subarray sum, return the current index as the pivot index. If all candidate pivot indices are checked and no pivot index exists, return -1.

Java

Java

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