Leetcode之Max Sum of Rectangle No Larger Than K

題目：

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:

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

代碼：

方法一——暴力搜索法：

class Solution {
public:
    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
        if (matrix.empty() || matrix[0].empty()) return 0;
        int m = matrix.size(), n = matrix[0].size(), res = INT_MIN;
        int sum[m][n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int t = matrix[i][j];
                if (i > 0) t += sum[i - 1][j];
                if (j > 0) t += sum[i][j - 1];
                if (i > 0 && j > 0) t -= sum[i - 1][j - 1];
                sum[i][j] = t;
                for (int r = 0; r <= i; ++r) {
                    for (int c = 0; c <= j; ++c) {
                        int d = sum[i][j];
                        if (r > 0) d -= sum[r - 1][j];
                        if (c > 0) d -= sum[i][c - 1];
                        if (r > 0 && c > 0) d += sum[r - 1][c - 1];
                        if (d <= k) res = max(res, d);
                    }
                }
            }
        }
        return res;
    }
};

會超時

方法二——基於求二維數組最大子矩陣和的算法

class Solution {
public:
    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
        if (matrix.empty() || matrix[0].empty()) return 0;
        int m = matrix.size(), n = matrix[0].size(), res = INT_MIN;
        for (int i = 0; i < n; ++i) {
            vector<int> sum(m);
            for (int j = i; j < n; ++j) {
                for (int k = 0; k < m; ++k) {
                    sum[k] += matrix[k][j];
                }
                int curSum = 0;
                set<int> st{{0}};
                for (auto a : sum) {
                    curSum += a;
                    auto it = st.lower_bound(curSum - k);
                    if (it != st.end()) res = max(res, curSum - *it);
                    st.insert(curSum);
                }
            }
        }
        return res;
    }
};

思路：

矩陣有n列，就有n個起始列，遍歷n個起始列；對於每次第i列作爲起始列，建立一個大小爲m的sum數組，然後對i到n列進行遍歷，每次遍歷得到一個sum數組(sum[k] += matrix[k][j])，這個數組是一維的，求這個一維數組的最大字段和，代表着求相應矩形的最大矩陣元素和。