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 thematrix 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;      }  }