# 二维区域和检索 

<p><big><small>给定一个二维矩阵 <code>matrix</code>，</small></big>以下类型的多个请求：</p>

<ul>
	<li><big><small>计算其子矩形范围内元素的总和，该子矩阵的 <strong>左上角</strong> 为 <code>(row1,&nbsp;col1)</code> ，<strong>右下角</strong> 为 <code>(row2,&nbsp;col2)</code> 。</small></big></li>
</ul>

<p>实现 <code>NumMatrix</code> 类：</p>

<ul>
	<li><code>NumMatrix(int[][] matrix)</code>&nbsp;给定整数矩阵 <code>matrix</code> 进行初始化</li>
	<li><code>int sumRegion(int row1, int col1, int row2, int col2)</code>&nbsp;返回<big><small> <strong>左上角</strong></small></big><big><small> <code>(row1,&nbsp;col1)</code>&nbsp;、<strong>右下角</strong>&nbsp;<code>(row2,&nbsp;col2)</code></small></big> 所描述的子矩阵的元素 <strong>总和</strong> 。</li>
</ul>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<p><img src="https://pic.leetcode-cn.com/1626332422-wUpUHT-image.png" style="width: 200px;" /></p>

<pre>
<strong>输入:</strong> 
["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]]
<strong>输出:</strong> 
[null, 8, 11, 12]

<strong>解释:</strong>
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 (红色矩形框的元素总和)
numMatrix.sumRegion(1, 1, 2, 2); // return 11 (绿色矩形框的元素总和)
numMatrix.sumRegion(1, 2, 2, 4); // return 12 (蓝色矩形框的元素总和)
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m,&nbsp;n &lt;=&nbsp;200</code><meta charset="UTF-8" /></li>
	<li><code>-10<sup>5</sup>&nbsp;&lt;= matrix[i][j] &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= row1 &lt;= row2 &lt; m</code></li>
	<li><code>0 &lt;= col1 &lt;= col2 &lt; n</code></li>
	<li><meta charset="UTF-8" />最多调用 <code>10<sup>4</sup></code> 次&nbsp;<code>sumRegion</code> 方法</li>
</ul>


## template

```cpp


```

## 答案

```cpp

```

## 选项

### A

```cpp

```

### B

```cpp

```

### C

```cpp

```