Submatrix Sum

 1 public class Solution {
 2     /**
 3      * @param matrix an integer matrix
 4      * @return the coordinate of the left-up and right-down number
 5      */
 6     public int[][] submatrixSum(int[][] matrix) {
 7         int[][] res = new int[2][2];
 8         if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return res;
 9         
10         int m = matrix.length;
11         int n = matrix[0].length;
12         // sum from 0,0 to i, j
13         int[][] sum = new int[m + 1][n + 1];
14         
15         for (int i = 1; i < sum.length; i++) {
16             for (int j = 1; j < sum[0].length; j++) {
17                 sum[i][j] = matrix[i - 1][j - 1] + sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
18             }
19         }
20         //如果那个为0的矩阵在row i 和 j，那么我们可以把从i到j的那部分矩阵从上到下加起来，这样组成了一个一维数组，然后用Subarray Sum的方法解就可以了
21         
22 
23         for (int r1 = 0; r1 < m; r1++) {
24             for (int r2 = r1 + 1; r2 <= m; r2++) {
25                 Map<Integer, Integer> map = new HashMap<Integer, Integer>();
26                 for (int j = 0; j <= n; j++) {
27                     int zeroToJSum = sum[r2][j] - sum[r1][j];
28                     if (map.containsKey(zeroToJSum)) {
29                         res[0][0] = r1;
30                         res[0][1] = map.get(zeroToJSum);
31                         res[1][0] = r2 - 1;
32                         res[1][1] = j - 1;
33                         return res;
34                     } else {
35                         map.put(zeroToJSum, j);
36                     }
37                 }
38             }
39         }
40         return res;
41     }
42 }