[Swift]LeetCode308. 二维区域和检索 - 可变 $ 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.

---

给定一个二维矩阵，找到的数目的元素内部定义的rectangle市ITS的左上角(row1, col1)和右下角(row2, col2).

上面的矩形（带红色边框）由（row1，col1）=（2，1）和（row2，col2）=（4，3）定义，其中包含sum=8。 

实例：

给定 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：

只读modifiable冰矩阵的更新功能。

你可以要求银行assume'号更新功能和sumregion evenly冰的分布。

你可能是 row1 ≤ row2 且 col1 ≤ col2.

---

Solution:

class NumMatrix {
    var mat:[[Int]] = [[Int]]()
    var colSum:[[Int]] = [[Int]]()
    init(_ matrix:inout [[Int]])
    {
        if matrix.isEmpty || matrix[0].isEmpty
        {
            return
        }
        mat = matrix
        colSum = [[Int]](repeating:[Int](repeating:0,count:matrix[0].count),count:matrix.count + 1)
        for i in 1..<colSum.count
        {
            for j in 0..<colSum[0].count
            {
                colSum[i][j] = colSum[i - 1][j] + matrix[i - 1][j]
            }
        }
    }
    
    func update(_ row:Int,_ col:Int,_ val:Int)
    {
        for i in (row + 1)..<colSum.count
        {
            colSum[i][col] += val - mat[row][col]
        }
        mat[row][col] = val
    }
    
    func sumRegion(_ row1:Int,_ col1:Int,_ row2:Int,_ col2:Int) -> Int
    {
        var res:Int = 0
        for j in col1...col2
        {
            res += colSum[row2 + 1][j] - colSum[row1][j]
        }
        return res
    }
}

点击：Playground测试

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]]
var sol = NumMatrix(&matrix)
print(sol.sumRegion(2, 1, 4, 3))
//Print 8
sol.update(3, 2, 2)
print(sol.sumRegion(2, 1, 4, 3))
//Print 10