mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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204 lines
29 KiB
JSON
204 lines
29 KiB
JSON
{
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"question": {
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"questionId": "914",
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"questionFrontendId": "497",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1793,
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"title": "Random Point in Non-overlapping Rectangles",
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"titleSlug": "random-point-in-non-overlapping-rectangles",
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"content": "<p>You are given an array of non-overlapping axis-aligned rectangles <code>rects</code> where <code>rects[i] = [a<sub>i</sub>, b<sub>i</sub>, x<sub>i</sub>, y<sub>i</sub>]</code> indicates that <code>(a<sub>i</sub>, b<sub>i</sub>)</code> is the bottom-left corner point of the <code>i<sup>th</sup></code> rectangle and <code>(x<sub>i</sub>, y<sub>i</sub>)</code> is the top-right corner point of the <code>i<sup>th</sup></code> rectangle. Design an algorithm to pick a random integer point inside the space covered by one of the given rectangles. A point on the perimeter of a rectangle is included in the space covered by the rectangle.</p>\n\n<p>Any integer point inside the space covered by one of the given rectangles should be equally likely to be returned.</p>\n\n<p><strong>Note</strong> that an integer point is a point that has integer coordinates.</p>\n\n<p>Implement the <code>Solution</code> class:</p>\n\n<ul>\n\t<li><code>Solution(int[][] rects)</code> Initializes the object with the given rectangles <code>rects</code>.</li>\n\t<li><code>int[] pick()</code> Returns a random integer point <code>[u, v]</code> inside the space covered by one of the given rectangles.</li>\n</ul>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/07/24/lc-pickrandomrec.jpg\" style=\"width: 419px; height: 539px;\" />\n<pre>\n<strong>Input</strong>\n["Solution", "pick", "pick", "pick", "pick", "pick"]\n[[[[-2, -2, 1, 1], [2, 2, 4, 6]]], [], [], [], [], []]\n<strong>Output</strong>\n[null, [1, -2], [1, -1], [-1, -2], [-2, -2], [0, 0]]\n\n<strong>Explanation</strong>\nSolution solution = new Solution([[-2, -2, 1, 1], [2, 2, 4, 6]]);\nsolution.pick(); // return [1, -2]\nsolution.pick(); // return [1, -1]\nsolution.pick(); // return [-1, -2]\nsolution.pick(); // return [-2, -2]\nsolution.pick(); // return [0, 0]\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= rects.length <= 100</code></li>\n\t<li><code>rects[i].length == 4</code></li>\n\t<li><code>-10<sup>9</sup> <= a<sub>i</sub> < x<sub>i</sub> <= 10<sup>9</sup></code></li>\n\t<li><code>-10<sup>9</sup> <= b<sub>i</sub> < y<sub>i</sub> <= 10<sup>9</sup></code></li>\n\t<li><code>x<sub>i</sub> - a<sub>i</sub> <= 2000</code></li>\n\t<li><code>y<sub>i</sub> - b<sub>i</sub> <= 2000</code></li>\n\t<li>All the rectangles do not overlap.</li>\n\t<li>At most <code>10<sup>4</sup></code> calls will be made to <code>pick</code>.</li>\n</ul>\n",
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"translatedTitle": "非重叠矩形中的随机点",
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"translatedContent": "<p>给定一个由非重叠的轴对齐矩形的数组 <code>rects</code> ,其中 <code>rects[i] = [ai, bi, xi, yi]</code> 表示 <code>(ai, bi)</code> 是第 <code>i</code> 个矩形的左下角点,<code>(xi, yi)</code> 是第 <code>i</code> 个矩形的右上角点。设计一个算法来随机挑选一个被某一矩形覆盖的整数点。矩形周长上的点也算做是被矩形覆盖。所有满足要求的点必须等概率被返回。</p>\n\n<p>在给定的矩形覆盖的空间内的任何整数点都有可能被返回。</p>\n\n<p><strong>请注意 </strong>,整数点是具有整数坐标的点。</p>\n\n<p>实现 <code>Solution</code> 类:</p>\n\n<ul>\n\t<li><code>Solution(int[][] rects)</code> 用给定的矩形数组 <code>rects</code> 初始化对象。</li>\n\t<li><code>int[] pick()</code> 返回一个随机的整数点 <code>[u, v]</code> 在给定的矩形所覆盖的空间内。</li>\n</ul>\n\n<ol>\n</ol>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<p><img src=\"https://assets.leetcode.com/uploads/2021/07/24/lc-pickrandomrec.jpg\" style=\"height: 539px; width: 419px;\" /></p>\n\n<pre>\n<strong>输入: \n</strong>[\"Solution\", \"pick\", \"pick\", \"pick\", \"pick\", \"pick\"]\n[[[[-2, -2, 1, 1], [2, 2, 4, 6]]], [], [], [], [], []]\n<strong>输出: \n</strong>[null, [1, -2], [1, -1], [-1, -2], [-2, -2], [0, 0]]\n\n<strong>解释:</strong>\nSolution solution = new Solution([[-2, -2, 1, 1], [2, 2, 4, 6]]);\nsolution.pick(); // 返回 [1, -2]\nsolution.pick(); // 返回 [1, -1]\nsolution.pick(); // 返回 [-1, -2]\nsolution.pick(); // 返回 [-2, -2]\nsolution.pick(); // 返回 [0, 0]</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 <= rects.length <= 100</code></li>\n\t<li><code>rects[i].length == 4</code></li>\n\t<li><code>-10<sup>9</sup> <= a<sub>i</sub> < x<sub>i</sub> <= 10<sup>9</sup></code></li>\n\t<li><code>-10<sup>9</sup> <= b<sub>i</sub> < y<sub>i</sub> <= 10<sup>9</sup></code></li>\n\t<li><code>x<sub>i</sub> - a<sub>i</sub> <= 2000</code></li>\n\t<li><code>y<sub>i</sub> - b<sub>i</sub> <= 2000</code></li>\n\t<li>所有的矩形不重叠。</li>\n\t<li><code>pick</code> 最多被调用 <code>10<sup>4</sup></code> 次。</li>\n</ul>\n",
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