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].

Code:

class Solution {
    public int pivotIndex(int[] nums) {
        int len = nums.length;
        int sum = 0, leftSum = 0;
        
        for (int i = 0; i < len; i++) {
            sum += nums[i]; //数组求和
        }
        
        for (int i = 0; i < len; i++) {
            if (i != 0) leftSum += nums[i - 1];
            if (sum - leftSum - nums[i] == leftSum) return i;
        }
        return -1;
    }
}

---

• 该题的解题思路是先求出数组的和，然后算左边的累加，如果数组和减去左边的累加等于左边的累加，则可以返回pivot。
• 注意pivot index在第一位的情况。