[leetcode]363. Max Sum of Rectangle No Larger Than K

最新推荐文章于 2019-04-07 11:36:07 发布

xiaocong1990

最新推荐文章于 2019-04-07 11:36:07 发布

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本文探讨了LeetCode上一道经典题目：寻找矩阵中不超过k的最大子矩阵和。通过动态规划的方法，文章详细介绍了如何有效地解决这个问题，并给出了具体的实现代码。

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题目链接：https://leetcode.com/problems/max-sum-of-sub-matrix-no-larger-than-k/#/description

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?

class Solution {
public:
    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) 
    {
        if(matrix.empty()) return 0;
        int rowSize = matrix.size(), colSize = matrix[0].size();
        int ret = INT_MIN;
        for(int l = 0; l < colSize; ++l) //starting leftmost column;
        {
            vector<int> sums(rowSize, 0); //store the row pre-sums;
            for(int c = l; c < colSize; ++c) //try different ending columns;
            {
                for(int r = 0; r < rowSize; ++r) //sum them up in rows;
                    sums[r] += matrix[r][c];
                set<int> sums_set; //store the sums from the starting top-left;
                sums_set.insert(0); //as a sentinel;
                int maxSum = INT_MIN, sum = 0;
                for(int i = 0; i < rowSize; ++i)
                {
                    sum += sums[i]; //the sum from the starting top-left to current position;
                    auto iter = sums_set.lower_bound(sum-k); //check the possible sum candidates;
                    if(iter != sums_set.end()) maxSum = max(maxSum, sum-*iter); //found one, check it;
                    sums_set.insert(sum);
                }
                ret = max(ret, maxSum);
            }
        }
        return ret;
    }
};