[LintCode] 1840. Matrix restoration

There is a matrix beforebefore with nn rows and mm columns. For each element in before before[i][j]before[i][j], we will use the following algorithm to convert it to after [i] [j]after[i][j]. Given the afterafter matrix, please restore the original matrix beforebefore.

s = 0
for i1: 0 -> i
    for j1: 0 -> j
        s = s + before[i1][j1]
after[i][j] = s

Example

Input:
2
2
[[1,3],[4,10]]
Output: [[1,2],[3,4]]
Explanation:
before:
1 2
3 4

after:
1 3
4 10

public class Solution {
    /**
     * @param n: the row of the matrix
     * @param m: the column of the matrix
     * @param after: the matrix
     * @return: restore the matrix
     */
    public int[][] matrixRestoration(int n, int m, int[][] after) {
        // write your code here
        // pre[i][j] = pre[i - 1][j] + pre[i][j - 1] - pre[i- 1][j - 1] + grid[i][j]
        // grid[i][j] = pre[i][j] - pre[i - 1][j] - pre[i][j - 1] + pre[i- 1][j - 1] 
        int[][] res = new int[after.length][after[0].length];
        for(int i = 0; i < after.length; i++) {
            for (int j = 0; j < after[0].length; j++) {
                res[i][j] = after[i][j];
                if (i > 0) {
                    res[i][j] -= after[i - 1][j];
                }
                if (j > 0) {
                    res[i][j] -= after[i][j - 1];
                }
                if (i > 0 && j > 0) {
                    res[i][j] += after[i - 1][j - 1];
                }
            }
        }
        return res;
    }
}