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leetcode-cn/originData/range-sum-query-2d-immutable.json
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@@ -12,7 +12,7 @@
"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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