304. 二维区域和检索 - 矩阵不可变

最新推荐文章于 2025-02-12 15:50:34 发布

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最新推荐文章于 2025-02-12 15:50:34 发布

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本文介绍了一种高效计算二维矩阵中任意矩形区域元素之和的方法。通过预处理生成积分图，实现快速查找指定矩形区域的元素总和，适用于矩阵不变且需频繁查询的情况。

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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).

Range Sum Query 2D
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
sumRegion(1, 1, 2, 2) -> 11
sumRegion(1, 2, 2, 4) -> 12

Note:

You may assume that the matrix does not change.
There are many calls to sumRegion function.
You may assume that row1 ≤ row2 and col1 ≤ col2.

Review:

以前做过类似的矩阵面积的题,init时先将二维矩阵转换为积分图

构造一个2D数组 sums[row+1][col+1]

（注意：我们添加了额外的空白行sums[0][col+1]={0}和空白列sums[row+1][0]={0}来删除边缘情况检查），因此，我们可以有以下定义

sums[i+1][j+1]表示区域的从总和matrix[0][0]至matrix[i][j]

要计算总和，想法如下

+-----+-+-------+     +--------+-----+     +-----+---------+     +-----+--------+
|     | |       |     |        |     |     |     |         |     |     |        |
|     | |       |     |        |     |     |     |         |     |     |        |
+-----+-+       |     +--------+     |     |     |         |     +-----+        |
|     | |       |  =  |              |  +  |     |         |  -  |              |
+-----+-+       |     |              |     +-----+         |     |              |
|               |     |              |     |               |     |              |
|               |     |              |     |               |     |              |
+---------------+     +--------------+     +---------------+     +--------------+

   sums[i][j]      =    sums[i-1][j]    +     sums[i][j-1]    -   sums[i-1][j-1]   +  

                        matrix[i-1][j-1]

So, we use the same idea to find the specific area's sum.

+---------------+   +--------------+   +---------------+   +--------------+   +--------------+
|               |   |         |    |   |   |           |   |         |    |   |   |          |
|   (r1,c1)     |   |         |    |   |   |           |   |         |    |   |   |          |
|   +------+    |   |         |    |   |   |           |   +---------+    |   +---+          |
|   |      |    | = |         |    | - |   |           | - |      (r1,c2) | + |   (r1,c1)    |
|   |      |    |   |         |    |   |   |           |   |              |   |              |
|   +------+    |   +---------+    |   +---+           |   |              |   |              |
|        (r2,c2)|   |       (r2,c2)|   |   (r2,c1)     |   |              |   |              |
+---------------+   +--------------+   +---------------+   +--------------+   +--------------+

Code:

python:

class NumMatrix:

    def __init__(self, matrix):
        self.mtx = matrix
        if self.mtx:
            dp = [[]]
            for i in range(1, len(matrix)+1):
                for j in range(1, len(matrix[0])+1):
                    dp[i][j] = matrix[i - 1][j - 1] + dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1]
            self._dp = dp

    def sumRegion(self, row1: int, col1: int, row2: int, col2: int) -> int:
        res = 0
        if self.mtx:
            res = self._dp[row2 + 1][col2 + 1] - self._dp[row2 + 1][col1] - self._dp[row1][col2 + 1] + self._dp[row1][
                col1]
        return res

java:

class NumMatrix {

    static int[][] mtx;
    static int[][] dp;
    public NumMatrix(final int[][] matrix) {
        mtx = matrix;
        if (mtx.length!=0){
            dp = new int[mtx.length+1][matrix[0].length+1];
            for (int i = 1; i < dp.length; i++) {
                for (int j = 1; j < dp[0].length; j++) {
                    dp[i][j] = matrix[i - 1][j - 1] + dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1];
                }
            }
        }
    }

    public int sumRegion(int row1, int col1, int row2, int col2) {
        int res = 0;
        if (mtx.length!=0)
            res = dp[row2+1][col2+1] - dp[row2+1][col1] - dp[row1][col2+1] + dp[row1][col1];
        return res;
    }
}