leetcode 363. Max Sum of Rectangle No Larger Than K

写在前面

这题被标记为hard，第一眼看到，确实很容易想到是dp，但思路之后就陷入混乱，原因就是不知道dp的状态转移方程，以及dp过程的开始和结束，本题事实上是两道题目的组合，分别是 Max Sum of Rectangle in a matrix,这道题的题解可以看这个视频（需要）
以及这道题目：max subarray sum no more than k
本题基本上是以上两种思路的组合了，具体的实现很简单。

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

解题思路

具体的思路在第一小节已经说明了，这里简单提一下第二道题目的题解，利用容器计算 sum(i,j)的技巧我们之前也多次遇到过，这种思路的题目会在后面再补充一道。

代码实现

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

        if(matrix.empty()) return 0;
        int rows = matrix.size();
        int cols = matrix[0].size();
        int best = INT_MIN;
        for(int left = 0;leftvector<int> sums(rows,0);
            for(int right = left;rightfor(int i = 0;iset<int> contain;
                contain.insert(0);
                int curr = 0;
                for(auto val:sums) {
                    curr+=val;
                    auto iter = contain.lower_bound(curr-k);
                    if(iter!=contain.end()) best = max(best,curr-*iter);
                    contain.insert(curr);
                }
            }
        }
        return best;

    }
};