Leetcode 1991. Find the middle position of the array (brute force enumeration)

I'll give you a subscript from 0 The starting array of integers nums , Please find Leftmost In the middle of middleIndex （ That is, the one with the smallest subscript in all possible middle positions ）.

In the middle middleIndex Is to meet nums[0] + nums[1] + … + nums[middleIndex-1] == nums[middleIndex+1] + nums[middleIndex+2] + … + nums[nums.length-1] Array subscript of .

If middleIndex == 0 , The sum of the left part is defined as 0 . Allied , If middleIndex == nums.length - 1 , The sum of the right part is defined as 0 .

Please return to meet the above conditions Leftmost Of middleIndex , If there is no such middle position , Please return -1 .

Example 1：

 Input ：nums = [2,3,-1,8,4]
 Output ：3
 explain ：
 Subscript  3  The sum of the previous numbers is ：2 + 3 + -1 = 4
 Subscript  3  The following numbers and are ：4 = 4

Example 2：

 Input ：nums = [1,-1,4]
 Output ：2
 explain ：
 Subscript  2  The sum of the previous numbers is ：1 + -1 = 0
 Subscript  2  The following numbers and are ：0

Example 3：

 Input ：nums = [2,5]
 Output ：-1
 explain ：
 There are no qualified  middleIndex .

Example 4：

 Input ：nums = [1]
 Output ：0
 explain ：
 Subscript  0  The sum of the previous numbers is ：0
 Subscript  0  The following numbers and are ：0

Tips ：

• 1 <= nums.length <= 100
• -1000 <= nums[i] <= 1000

Code:

class Solution {
    
public:
    int findMiddleIndex(vector<int>& nums) {
    
        int left=0;
        int right=0;
        for(int i=0;i<nums.size();i++)
        {
    
            if(i==0)
            {
    
                left=0;
                right=accumulate(nums.begin()+i+1,nums.end(),0);
                if(left==right)
                    return i;
            }
            if(i==nums.size()-1)
            {
    
                right=0;
                left=accumulate(nums.begin(),nums.end()-1,0);
                if(left==right)
                    return i;
            }
            left=accumulate(nums.begin(),nums.begin()+i,0);
            right=accumulate(nums.begin()+i+1,nums.end(),0);
            if(left==right)
                return i;
        }
        return -1;
    }
};