[Swift]LeetCode363. 矩形区域不超过 K 的最大数值和 | Max Sum of Rectangle No Larger Than K

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 = [[1,1],[0,-2,3]],k = 2 Output: 2 Explanation: Because the sum of rectangle  is 2,  and 2 is the max number no larger than k (k = 2).[[0,[-2,3]]

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

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

 1 class Solution {
 2     func maxSumSubmatrix(_ matrix: [[Int]],_ k: Int) -> Int {
 3         if matrix.isEmpty || matrix[0].isEmpty {return 0}
 4         var m:Int = matrix.count
 5         var n:Int = matrix[0].count
 6         var res:Int = Int.min
 7         var sum:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:n),count:m)
 8         for i in 0..<m
 9         {
10             for j in 0..<n
11             {
12                 var t:Int = matrix[i][j]
13                 if i > 0 {t += sum[i - 1][j]}
14                 if j > 0 {t += sum[i][j - 1]}
15                 if i > 0 && j > 0 {t -= sum[i - 1][j - 1]}
16                 sum[i][j] = t
17                 for r in 0...i
18                 {
19                     for c in 0...j
20                     {
21                         var d:Int = sum[i][j]
22                         if r > 0 {d -= sum[r - 1][j]}
23                         if c > 0 {d -= sum[i][c - 1]}
24                         if r > 0 && c > 0 {d += sum[r - 1][c - 1]}
25                         if d <= k {res = max(res,d)}
26                     }
27                 }
28             }
29         }
30         return res
31     }
32 }