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Java-based LeetCode algorithm problem solutions, regularly updated
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304\. Range Sum Query 2D - Immutable
Medium
Given a 2D matrix `matrix`, handle multiple queries of the following type:
* Calculate the **sum** of the elements of `matrix` inside the rectangle defined by its **upper left corner** `(row1, col1)` and **lower right corner** `(row2, col2)`.
Implement the NumMatrix class:
* `NumMatrix(int[][] matrix)` Initializes the object with the integer matrix `matrix`.
* `int sumRegion(int row1, int col1, int row2, int col2)` Returns the **sum** of the elements of `matrix` inside the rectangle defined by its **upper left corner** `(row1, col1)` and **lower right corner** `(row2, col2)`.
**Example 1:**
![](https://assets.leetcode.com/uploads/2021/03/14/sum-grid.jpg)
**Input**
["NumMatrix", "sumRegion", "sumRegion", "sumRegion"]
[[[[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, 1, 4, 3], [1, 1, 2, 2], [1, 2, 2, 4]]
**Output:** [null, 8, 11, 12]
**Explanation:**
NumMatrix numMatrix = new NumMatrix([[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]]);
numMatrix.sumRegion(2, 1, 4, 3); // return 8 (i.e sum of the red rectangle)
numMatrix.sumRegion(1, 1, 2, 2); // return 11 (i.e sum of the green rectangle)
numMatrix.sumRegion(1, 2, 2, 4); // return 12 (i.e sum of the blue rectangle)
**Constraints:**
* `m == matrix.length`
* `n == matrix[i].length`
* `1 <= m, n <= 200`
* -105 <= matrix[i][j] <= 105
* `0 <= row1 <= row2 < m`
* `0 <= col1 <= col2 < n`
* At most 104
calls will be made to `sumRegion`.
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