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47 lines
2.1 KiB
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47 lines
2.1 KiB
HTML
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<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>
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<ul>
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<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, col1)</code> ，<strong>右下角</strong> 为 <code>(row2, col2)</code> 。</small></big></li>
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</ul>
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<p>实现 <code>NumMatrix</code> 类：</p>
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<ul>
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<li><code>NumMatrix(int[][] matrix)</code> 给定整数矩阵 <code>matrix</code> 进行初始化</li>
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<li><code>int sumRegion(int row1, int col1, int row2, int col2)</code> 返回<big><small> <strong>左上角</strong></small></big><big><small> <code>(row1, col1)</code> 、<strong>右下角</strong> <code>(row2, col2)</code></small></big> 所描述的子矩阵的元素 <strong>总和</strong> 。</li>
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</ul>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<p><img src="https://pic.leetcode-cn.com/1626332422-wUpUHT-image.png" style="width: 200px;" /></p>
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<pre>
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<strong>输入:</strong>
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["NumMatrix","sumRegion","sumRegion","sumRegion"]
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[[[[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]]
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<strong>输出:</strong>
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[null, 8, 11, 12]
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<strong>解释:</strong>
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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]]);
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numMatrix.sumRegion(2, 1, 4, 3); // return 8 (红色矩形框的元素总和)
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numMatrix.sumRegion(1, 1, 2, 2); // return 11 (绿色矩形框的元素总和)
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numMatrix.sumRegion(1, 2, 2, 4); // return 12 (蓝色矩形框的元素总和)
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>m == matrix.length</code></li>
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<li><code>n == matrix[i].length</code></li>
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<li><code>1 <= m, n <= 200</code><meta charset="UTF-8" /></li>
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<li><code>-10<sup>5</sup> <= matrix[i][j] <= 10<sup>5</sup></code></li>
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<li><code>0 <= row1 <= row2 < m</code></li>
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<li><code>0 <= col1 <= col2 < n</code></li>
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<li><meta charset="UTF-8" />最多调用 <code>10<sup>4</sup></code> 次 <code>sumRegion</code> 方法</li>
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</ul>
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