【leetcode】363. Max Sum of Rectangle No Larger Than K

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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:

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?

假设矩阵行总数为row，列总数为col，设置矩形的起始行为i，结束行为j，创建row个存储矩形总和的数组，并依次进行遍历迭代。

public class Solution {
    public int maxSumSubmatrix(int[][] matrix, int k) {
        int row = matrix.length;
        int col = matrix[0].length;
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < row; i ++) {
            int[] colSum = new int[col];
            for (int j = i; j < row; j ++) {
                for (int c = 0; c < col; c ++) {
                    colSum[c] += matrix[j][c];
                }
                max = Math.max(max, findMax(colSum, k));
            }
        }
        return max;
    }
    
    private int findMax(int[] nums, int k) {
        int max = Integer.MIN_VALUE;
        int sum = 0;
        TreeSet<Integer> s = new TreeSet();
        s.add(0);
    
        for(int i = 0;i < nums.length; i ++){
            int t = sum + nums[i];
            sum = t;
            Integer gap = s.ceiling(sum - k);
            if(gap != null) max = Math.max(max, sum - gap);
            s.add(t);
        }
    
        return max;
    }
}