Maximum Submatrix - LintCode

Given an n x n matrix of positive and negative integers, find the submatrix with the largest possible sum.

Sample

Given matrix = 
[
[1,3,-1],
[2,3,-2],
[-1,-2,-3]
]
return 9.
Explanation:
the submatrix with the largest possible sum is:
[
[1,2],
[2,3]
]

The idea is to
traverse the array and set the value of matrix[i][j] to the maximum value of the elements of the matrix with matrix[i][j] as the lower right corner.

#ifndef C944_H
#define C944_H
#include<iostream>
#include<vector>
using namespace std;
class Solution {
public:
    /**
    * @param matrix: the given matrix
    * @return: the largest possible sum
    */
    int maxSubmatrix(vector<vector<int>> &matrix) {
        // write your code here
        if (matrix.empty() || matrix[0].empty())
            return 0;
        int res = 0;
        int rows = matrix.size();
        int cols = matrix[0].size();
        vector<vector<int>> sum = matrix;
        //sum[i][j]表示以matrix[0][0]为左上角,matrix[i][j]为右下角的矩阵,其包含元素的和
        for (int i = 1; i < rows; ++i)
            sum[i][0] += sum[i - 1][0];
        for (int j = 1; j < cols; ++j)
            sum[0][j] += sum[0][j - 1];
        for (int i = 1; i < rows; ++i)
        {
            for (int j = 1; j < cols; ++j)
            {
                sum[i][j] = sum[i][j] + sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
            }
        }
        //遍历数组,将matrix[i][j]的值置为以matrix[i][j]为右下角的矩阵其元素的最大值
        for (int i = 0; i < rows; ++i)
        {
            for (int j = 0; j < cols; ++j)
            {
                int val = INT_MIN;
                for (int k = 0; k <= i; ++k)
                {
                    for (int l = 0; l <= j; ++l)
                    {
                        int temp = 0;
                        if (k == 0 && l>0)
                            temp = sum[i][j] - sum[i][l - 1];
                        else if (k > 0 && l == 0)
                            temp = sum[i][j] - sum[k - 1][j];
                        else if (k == 0 && l == 0)
                            temp = sum[i][j];
                        else
                            temp = sum[i][j] - sum[k - 1][j] - sum[i][l - 1] + sum[k - 1][l - 1];
                        val = maxVal(val, temp);
                    }
                }
                matrix[i][j] = val;
            }
        }
        for (auto c : matrix)
        {
            for (auto t : c)
            {
                res = maxVal(t, res);
            }
        }
        return res;
    }
    int maxVal(int a, int b)
    {
        return a > b ? a : b;
    }
};
#endif