[Swift]LeetCode308. 二維區域和檢索 - 可變 $ Range Sum Query 2D - Mutable

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Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.

Example:

Given matrix = [
  [3, 0, 1, 4, 2],
  [5, 6, 3, 2, 1],
  [1, 2, 0, 1, 5],
  [4, 1, 0, 1, 7],
  [1, 0, 3, 0, 5]
]

sumRegion(2, 1, 4, 3) -> 8
update(3, 2, 2)
sumRegion(2, 1, 4, 3) -> 10

Note:

1. The matrix is only modifiable by the update function.
2. You may assume the number of calls to update and sumRegion function is distributed evenly.
3. You may assume that row1 ≤ row2 and col1 ≤ col2.

---

給定一個二維矩陣，找到的數目的元素內部定義的rectangle市ITS的左上角(row1, col1)和右下角(row2, col2).

上面的矩形（帶紅色邊框）由（row1，col1）=（2，1）和（row2，col2）=（4，3）定義，其中包含sum=8。 

實例：

給定 matrix = [
  [3, 0, 1, 4, 2],
  [5, 6, 3, 2, 1],
  [1, 2, 0, 1, 5],
  [4, 1, 0, 1, 7],
  [1, 0, 3, 0, 5]
]

sumRegion(2, 1, 4, 3) -> 8
update(3, 2, 2)
sumRegion(2, 1, 4, 3) -> 10

注：

只讀modifiable冰矩陣的更新功能。

你能夠要求銀行assume'號更新功能和sumregion evenly冰的分佈。

你多是 row1 ≤ row2 且 col1 ≤ col2.

---

Solution:

 1 class NumMatrix {
 2     var mat:[[Int]] = [[Int]]()
 3     var colSum:[[Int]] = [[Int]]()
 4     init(_ matrix:inout [[Int]])
 5     {
 6         if matrix.isEmpty || matrix[0].isEmpty
 7         {
 8             return
 9         }
10         mat = matrix
11         colSum = [[Int]](repeating:[Int](repeating:0,count:matrix[0].count),count:matrix.count + 1)
12         for i in 1..<colSum.count
13         {
14             for j in 0..<colSum[0].count
15             {
16                 colSum[i][j] = colSum[i - 1][j] + matrix[i - 1][j]
17             }
18         }
19     }
20     
21     func update(_ row:Int,_ col:Int,_ val:Int)
22     {
23         for i in (row + 1)..<colSum.count
24         {
25             colSum[i][col] += val - mat[row][col]
26         }
27         mat[row][col] = val
28     }
29     
30     func sumRegion(_ row1:Int,_ col1:Int,_ row2:Int,_ col2:Int) -> Int
31     {
32         var res:Int = 0
33         for j in col1...col2
34         {
35             res += colSum[row2 + 1][j] - colSum[row1][j]
36         }
37         return res
38     }
39 }

點擊：Playground測試

1 var matrix:[[Int]] = [[3, 0, 1, 4, 2],[5, 6, 3, 2, 1],[1, 2, 0, 1, 5],[4, 1, 0, 1, 7],[1, 0, 3, 0, 5]]
2 var sol = NumMatrix(&matrix)
3 print(sol.sumRegion(2, 1, 4, 3))
4 //Print 8
5 sol.update(3, 2, 2)
6 print(sol.sumRegion(2, 1, 4, 3))
7 //Print 10