[leetcode]304. Range Sum Query 2D - Immutable

最新推荐文章于 2025-02-03 15:37:10 发布

xiaocong1990

最新推荐文章于 2025-02-03 15:37:10 发布

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本文介绍了一种高效解决二维矩阵区间求和问题的方法。通过预处理构建一个辅助矩阵，可以在常数时间内计算出任意矩形区域内的元素总和。文章详细解释了构建辅助矩阵的过程及其在求解区间和的应用。

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题目链接：https://leetcode.com/problems/range-sum-query-2d-immutable/#/description

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

Construct a 2D array sums[row+1][col+1]

(notice: we add additional blank row sums[0][col+1]={0} and blank column sums[row+1][0]={0} to remove the edge case checking), so, we can have the following definition

sums[i+1][j+1] represents the sum of area from matrix[0][0] to matrix[i][j]

To calculate sums, the ideas as below

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

   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)     |   |              |   |              |
+---------------+   +--------------+   +---------------+   +--------------+   +--------------+

class NumMatrix {
private:
    int row, col;
    vector<vector<int>> sums;
public:
    NumMatrix(vector<vector<int>> &matrix) {
        row = matrix.size();
        col = row>0 ? matrix[0].size() : 0;
        sums = vector<vector<int>>(row+1, vector<int>(col+1, 0));
        for(int i=1; i<=row; i++) {
            for(int j=1; j<=col; j++) {
                sums[i][j] = matrix[i-1][j-1] + 
                             sums[i-1][j] + sums[i][j-1] - sums[i-1][j-1] ;
            }
        }
    }

    int sumRegion(int row1, int col1, int row2, int col2) {
        return sums[row2+1][col2+1] - sums[row2+1][col1] - sums[row1][col2+1] + sums[row1][col1];
    }
};