leetcode 363. Max Sum of Rectangle No Larger Than K 子矩形和小于K的最大值 + 动态规划DP + 暴力循环真好

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?

没有想到更好地做法，我这里直接暴力求解。效果还不错，accept了。

代码如下：

import java.util.TreeSet;


class Solution 
{
    /*
     * 暴力做法很定不好，但是暂时想不到更好的办法，所以先这么办吧
     * */
    public int maxSumSubmatrix(int[][] matrix, int k) 
    {
        if(matrix==null || matrix.length<=0)
            return 0;

        int[][] sum=new int[matrix.length+1][matrix[0].length+1];       
        for(int i=1;i<=matrix.length;i++)
            for(int j=1;j<=matrix[0].length;j++)
                sum[i][j]=matrix[i-1][j-1]+sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];

        int max=Integer.MIN_VALUE;
        for(int i=1;i<=matrix.length;i++)
        {
            for(int j=1;j<=matrix[0].length;j++)
            {
                for(int g=0;gfor(int h=0;hint one=sum[i][j]-(sum[i][h]+sum[g][j]-sum[g][h]);
                        if(one<=k)
                            max = Math.max(max, one);
                    }
                }
            }
        }
        return max;
    }
}

下面就是C++的做法

就是一个DP来求解矩形面积，然后遍历暴力求解即可，

代码如下：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;



class Solution 
{
public:
    int maxSumSubmatrix(vector<vector<int>>& mat, int k)
    {
        if (mat.size() <= 0)
            return 0;
        int row = mat.size(), col = mat[0].size();
        vector<vector<int>> dp(row + 1, vector<int>(col + 1, 0));
        for (int i = 1; i <= row; i++)
        {
            for (int j = 1; j <= col; j++)
            {
                dp[i][j] = mat[i-1][j-1] + dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1];
            }
        }

        int res = INT_MIN;
        for (int i = 1; i <= row; i++)
        {
            for (int j = 1; j <= col; j++)
            {
                for (int ii = 0; ii < i; ii++)
                {
                    for (int jj = 0; jj < j; jj++)
                    {
                        int one = dp[i][j] - dp[ii][j] - dp[i][jj] + dp[ii][jj];
                        if(one<=k)
                            res = max(res, one);
                    }
                }
            }
        }
        return res;
    }
};