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（需要翻墙）
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;left<cols;++left) {            vector<int> sums(rows,0);            for(int right = left;right<cols;++right) {                for(int i = 0;i<rows;++i) {                    sums[i]+= matrix[i][right];                }                set<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;    }};