[Swift]LeetCode363. 矩形区域不超过 K 的最大数值和 | 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?

---

给定一个非空二维矩阵 matrix 和一个整数 k，找到这个矩阵内部不大于 k的最大矩形和。

示例:

输入: matrix = [[1,0,1],[0,-2,3]], k = 2
输出: 2 
解释: 矩形区域 [[0, 1], [-2, 3]] 的数值和是 2，且 2 是不超过 k 的最大数字（k = 2）。

说明：

1. 矩阵内的矩形区域面积必须大于 0。
2. 如果行数远大于列数，你将如何解答呢？

---

13108ms

class Solution {
    func maxSumSubmatrix(_ matrix: [[Int]], _ k: Int) -> Int {
        if matrix.isEmpty || matrix[0].isEmpty {return 0}
        var m:Int = matrix.count
        var n:Int = matrix[0].count
        var res:Int = Int.min
        var sum:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:n),count:m)
        for i in 0..<m
        {
            for j in 0..<n
            {
                var t:Int = 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 r in 0...i
                {
                    for c in 0...j
                    {
                        var d:Int = 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
    }
}