Approach

# 切開左右兩邊的子陣列，相減計算是否為偶數
Split the array into left and right subarrays and check whether their difference is even.

Time Complexity

O(n * n)

Space Complexity

O(1)

Code

class Solution {
public:
    int countPartitions(vector<int>& nums) {
        int cnt = 0;

        for (int i = 1 ; i < nums.size() ; i++) {
            int sum1 = 0, sum2 = 0;

            for (int j = 0 ; j < i ; j++) {
                sum1 += nums[j];
            }

            for (int j = i ; j < nums.size() ; j++) {
                sum2 += nums[j];
            }

            //cout << sum1 << " " << sum2 << endl;
            if ((sum1 - sum2) % 2 == 0) {
                cnt++;
            }
        }

        return cnt;
    }
};