Max Sum of Rectangle No Larger Than K

Max Sum of Rectangle No Larger Than K

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:

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

解析：
使用累加求和来求每个子矩阵的和。

代码：

class Solution {
public:
int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {

int rows=matrix.size();
int cols=matrix[0].size();
vector<vector<int>> Sum(rows+1,vector<int>(cols+1,0));
int ans=INT_MIN;
for (int i=1; i<=rows; i++)
{
for (int j=1; j<=cols; j++)
{
Sum[i][j]=Sum[i-1][j]+Sum[i][j-1]-Sum[i-1][j-1]+matrix[i-1][j-1];

for (int r=0; r<i; r++)
{
for (int c=0; c<j; c++)
{
int temp=Sum[i][j]-Sum[r][j]-Sum[i][c]+Sum[r][c];
if (temp<=k&&temp>ans)
ans=temp;
}
}
}
}
return ans;

}
};