leetcode 363. Max Sum of Rectangle No Larger Than K_leetcode java 363. max sum of rectangle no larger -优快云博客

本文链接：https://blog.youkuaiyun.com/cmy203/article/details/89067114

本文介绍了一种寻找矩阵中不超过给定整数k的最大矩形和的算法。通过遍历矩阵的所有可能列组合，并使用集合维护当前行和，实现高效求解。示例输入为矩阵[[1,0,1],[0,-2,3]]与k=2，输出最大矩形和为2。

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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:

Input: matrix = [[1,0,1],[0,-2,3]], k = 2
Output: 2 
Explanation: 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

先占个坑，等我有空来写注释

class Solution {
public:
    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) 
    {
        int rowl = matrix.size(); //整个矩阵的行数
        int coll = matrix[0].size();//整个矩阵的列数
        int curmax;//curmax指的是ij列所有和小于K值的长方形的最大值
        int ans = INT_MIN;//ans指的是整个矩阵所有小于k值的长方形的最大值
        
        for(int i=0;i<coll;i++)
        {
            vector<int> sum(rowl,0);//记录ij列的每一行的和
            for(int j=i;j<coll;j++)
            {
                int cursum=0;//cursum指的是ij列(0,row)组成的长方形的和
                curmax = INT_MIN;
                set<int> st;//存放所有cursum值的集合
                st.insert(0);
                set<int>::iterator it;
                for(int row=0;row<rowl;row++)
                {
                    sum[row] += matrix[row][j];
                    cursum += sum[row];
                    it = st.lower_bound(cursum-k);
                    if(it != st.end())
                        curmax = max(curmax,cursum-*it);       
                    st.insert(cursum);
                }
                ans = max(ans,curmax);
            }
        }
        return ans;
    }
};