Count Numbers with Unique Digits

题目描述：

Given a non-empty 2D matrix matrix and an integer k, find the max sum of a rectangle in the matrix such that its sum is no larger than k.

Example:

Given matrix = [
  [1,  0, 1],
  [0, -2, 3]
]
k = 2

The answer is 2. Because the sum of rectangle [[0, 1], [-2, 3]] is 2 and 2 is the max number no larger than k (k = 2).

Note:

1. The rectangle inside the matrix must have an area > 0.
2. What if the number of rows is much larger than the number of columns?

自己最原始的想法是动态规划，看dp[i,j]和dp[i,j-1],dp[i,j-1],dp[i-1,j-1]之间的关系。

但是其实这个题只要把每个矩形的面积求出来就行了。

用sums[i][j]表示从( 0,0 )直到（i,j）范围内的矩阵和。

代码如下：

public class Solution {
    public int maxSumSubmatrix(int[][] matrix, int k) {  
        int row=matrix.length;  
        int col=matrix[0].length;  
        int[][] sums=new int[row][col];  
        sums[0][0]=matrix[0][0];  
        for (int i = 1; i < row; i++) {  
            sums[i][0]=sums[i-1][0]+matrix[i][0];  
        }  
        for (int j = 1; j < col; j++) {  
            sums[0][j]=sums[0][j-1]+matrix[0][j];  
        }  
        for (int i = 1; i <row; i++) {  
            for (int j = 1; j < col; j++) {  
                sums[i][j]=sums[i-1][j]+sums[i][j-1]-sums[i-1][j-1]+matrix[i][j];  
            }  
        }  
        int ans=Integer.MIN_VALUE;  
        for (int si = 0; si < row; si++) {  
            for (int sj = 0; sj < col; sj++) {  
                for (int ei = si; ei < row; ei++) {  
                    for (int ej = sj; ej < col; ej++) {  
                        int test=0;  
                        if (si==0&&sj==0) {  
                            test=sums[ei][ej];  
                        }else if(si==0){  
                            test=sums[ei][ej]-sums[ei][sj-1];  
                        }else if(sj==0){  
                            test=sums[ei][ej]-sums[si-1][ej];  
                        }else{  
                            test=sums[ei][ej]-sums[si-1][ej]-sums[ei][sj-1]+sums[si-1][sj-1];  
                        }  
                        if (test<=k&&test>ans) {  
                            ans=test;  
                        }  
                    }  
                }  
            }  
        }  
        return ans;  
    }  
}