leetcode-problemset/leetcode-cn/originData/range-sum-query-2d-immutable.json

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        "question": {
            "questionId": "304",
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            "categoryTitle": "Algorithms",
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            "title": "Range Sum Query 2D - Immutable",
            "titleSlug": "range-sum-query-2d-immutable",
            "content": "<p>Given a 2D matrix <code>matrix</code>, handle multiple queries of the following type:</p>\n\n<ul>\n\t<li>Calculate the <strong>sum</strong> of the elements of <code>matrix</code> inside the rectangle defined by its <strong>upper left corner</strong> <code>(row1, col1)</code> and <strong>lower right corner</strong> <code>(row2, col2)</code>.</li>\n</ul>\n\n<p>Implement the <code>NumMatrix</code> class:</p>\n\n<ul>\n\t<li><code>NumMatrix(int[][] matrix)</code> Initializes the object with the integer matrix <code>matrix</code>.</li>\n\t<li><code>int sumRegion(int row1, int col1, int row2, int col2)</code> Returns the <strong>sum</strong> of the elements of <code>matrix</code> inside the rectangle defined by its <strong>upper left corner</strong> <code>(row1, col1)</code> and <strong>lower right corner</strong> <code>(row2, col2)</code>.</li>\n</ul>\n\n<p>You must design an algorithm where <code>sumRegion</code> works on <code>O(1)</code> time complexity.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/03/14/sum-grid.jpg\" style=\"width: 415px; height: 415px;\" />\n<pre>\n<strong>Input</strong>\n[&quot;NumMatrix&quot;, &quot;sumRegion&quot;, &quot;sumRegion&quot;, &quot;sumRegion&quot;]\n[[[[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]]\n<strong>Output</strong>\n[null, 8, 11, 12]\n\n<strong>Explanation</strong>\nNumMatrix 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]]);\nnumMatrix.sumRegion(2, 1, 4, 3); // return 8 (i.e sum of the red rectangle)\nnumMatrix.sumRegion(1, 1, 2, 2); // return 11 (i.e sum of the green rectangle)\nnumMatrix.sumRegion(1, 2, 2, 4); // return 12 (i.e sum of the blue rectangle)\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>m == matrix.length</code></li>\n\t<li><code>n == matrix[i].length</code></li>\n\t<li><code>1 &lt;= m, n &lt;= 200</code></li>\n\t<li><code>-10<sup>4</sup> &lt;= matrix[i][j] &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= row1 &lt;= row2 &lt; m</code></li>\n\t<li><code>0 &lt;= col1 &lt;= col2 &lt; n</code></li>\n\t<li>At most <code>10<sup>4</sup></code> calls will be made to <code>sumRegion</code>.</li>\n</ul>\n",
            "translatedTitle": "二维区域和检索 - 矩阵不可变",
            "translatedContent": "<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>\n\n<ul>\n\t<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>\n</ul>\n\n<p>实现 <code>NumMatrix</code> 类：</p>\n\n<ul>\n\t<li><code>NumMatrix(int[][] matrix)</code>&nbsp;给定整数矩阵 <code>matrix</code> 进行初始化</li>\n\t<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>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img src=\"https://pic.leetcode-cn.com/1626332422-wUpUHT-image.png\" style=\"width: 200px;\" /></p>\n\n<pre>\n<strong>输入:</strong> \n[\"NumMatrix\",\"sumRegion\",\"sumRegion\",\"sumRegion\"]\n[[[[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]]\n<strong>输出:</strong> \n[null, 8, 11, 12]\n\n<strong>解释:</strong>\nNumMatrix 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]]);\nnumMatrix.sumRegion(2, 1, 4, 3); // return 8 (红色矩形框的元素总和)\nnumMatrix.sumRegion(1, 1, 2, 2); // return 11 (绿色矩形框的元素总和)\nnumMatrix.sumRegion(1, 2, 2, 4); // return 12 (蓝色矩形框的元素总和)\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>m == matrix.length</code></li>\n\t<li><code>n == matrix[i].length</code></li>\n\t<li><code>1 &lt;= m,&nbsp;n &lt;=&nbsp;200</code><meta charset=\"UTF-8\" /></li>\n\t<li><code>-10<sup>5</sup>&nbsp;&lt;= matrix[i][j] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= row1 &lt;= row2 &lt; m</code></li>\n\t<li><code>0 &lt;= col1 &lt;= col2 &lt; n</code></li>\n\t<li><meta charset=\"UTF-8\" />最多调用 <code>10<sup>4</sup></code> 次&nbsp;<code>sumRegion</code> 方法</li>\n</ul>\n",
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