724. Find Pivot Index

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].

My Solutions:

先循环一遍，把所有数字加起来，得到sum。再遍历一遍，sum减去当前数字，用一个left变量记录左半边的数字。如果对当前数字，left和sum相等则找到了这个数字。不然的话left加上当前数字，继续循环

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