Cells with Odd Values in a Matrix

给一个2d数组, 里面是横竖的index, 求在index上+1后matrix的奇数个数. 这个题我看有O(n)解法, 我自己做的是n^2

class Solution {
    public int oddCells(int n, int m, int[][] indices) {
        int[] row = new int[n];
        int[] col = new int[m];
        for(int[] i : indices) {
            row[i[0]]++;
            col[i[1]]++;
        }
        int count = 0;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++){
                if((row[i]+col[j]) % 2 != 0)
                    count++;
            }
        }
        return count;
    }
}

class Solution {
    public int oddCells(int n, int m, int[][] indices) {
        int[] row = new int[n];
        int[] col = new int[m];
        for(int[] i : indices) {
            row[i[0]]++;
            col[i[1]]++;
        }
        int count = 0;
        for(int i = 0; i < n; i++) {
            if(row[i] % 2 != 0) // if odd
                count++;
        }
        int res = 0;
        for(int i = 0; i < m; i++) {
            if(col[i] % 2 != 0) // if odd
                res += (n - count); // odd + odd = even
            else
                res += count;// odd + even = odd
        }
        return res;
    }
}