Submatrix Sum

Given an integer matrix, find a submatrix where the sum of numbers is zero. Your code should return the coordinate of the left-up and right-down number.

Example

Given matrix

[
  [1 ,5 ,7],
  [3 ,7 ,-8],
  [4 ,-8 ,9],
]

return [(1,1), (2,2)]

分析：

本质上还是subarray sum. 因为对于要找的那个submatrix, 一定在0 和 matrix.length 之间。假设那个submatrix的上下row分别为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。

 public class Solution {
     /**
      * @param matrix an integer matrix
      * @return the coordinate of the left-up and right-down number
      */
     public int[][] submatrixSum(int[][] matrix) {
         int[][] res = new int[][];
         if (matrix == null || matrix.length ==  || matrix[].length == ) return res;

         int m = matrix.length;
         int n = matrix[].length;
         // sum from 0,0 to i, j
         int[][] sum = new int[m + ][n + ];

         for (int i = ; i < sum.length; i++) {
             for (int j = ; j < sum[].length; j++) {
                 sum[i][j] = matrix[i - ][j - ] + sum[i - ][j] + sum[i][j - ] - sum[i - ][j - ];
             }
         }
         //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

         for (int r1 = ; r1 < m; r1++) {
             for (int r2 = r1 + ; r2 <= m; r2++) {
                 Map<Integer, Integer> map = new HashMap<Integer, Integer>();
                 for (int j = ; j <= n; j++) {
                     int zeroToJSum = sum[r2][j] - sum[r1][j];
                     if (map.containsKey(zeroToJSum)) {
                         res[][] = r1;
                         res[][] = map.get(zeroToJSum);
                         res[][] = r2 - ;
                         res[][] = j - ;
                         return res;
                     } else {
                         map.put(zeroToJSum, j);
                     }
                 }
             }
         }
         return res;
     }
 }

下面的代码思路和上面一样，只是实现不一样。

 public class Solution {
     /**
      * @param matrix an integer matrix
      * @return the coordinate of the left-up and right-down number
      */
     public int[][] submatrixSum(int[][] matrix) {
         int[][] res = new int[][];
         if (matrix == null || matrix.length ==  || matrix[].length == ) return res;

         int m = matrix.length;
         int n = matrix[].length;
         int[][] sum = new int[m][n];

         for (int i = ; i < m; i++) {
             for (int j = ; j < n; j++) {
                 if (i == ) {
                     sum[i][j] = matrix[i][j];
                 } else {
                     sum[i][j] = matrix[i][j] + sum[i - ][j];
                 }
             }
         }

         for (int r1 = ; r1 < m; r1++) {
             for (int r2 = r1; r2 < m; r2++) {
                 res = check(sum, r1, r2);
                 if (res != null) return res;
             }
         }
         return res;
     }

     private int[][] check(int[][] sum, int r1, int r2) {
         Map<Integer, Integer> map = new HashMap<Integer, Integer>();

         map.put(, -);
         int zeroToJSum = ;
         for (int j = ; j < sum[].length; j++) {
             if (r1 == ) {
                 zeroToJSum = sum[r2][j] + zeroToJSum;
             } else {
                 zeroToJSum = sum[r2][j] - sum[r1 - ][j] + zeroToJSum;
             }

             if (map.containsKey(zeroToJSum)) {
                 int[][] res = new int[][];
                 res[][] = r1;
                 res[][] = map.get(zeroToJSum) + ;
                 res[][] = r2;
                 res[][] = j;
                 return res;
             } else {
                 map.put(zeroToJSum, j);
             }
         }
         return null;
     }
 }

Reference:

https://segmentfault.com/a/1190000004878083