497:非重叠矩形中的随机点
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轩辕龙儿
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//<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>
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//
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//<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>
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//
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//<p><strong>请注意 </strong>，整数点是具有整数坐标的点。</p>
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//
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//<p>实现 <code>Solution</code> 类:</p>
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//
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//<ul>
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// <li><code>Solution(int[][] rects)</code> 用给定的矩形数组 <code>rects</code> 初始化对象。</li>
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// <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>
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//</ul>
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//
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//<ol>
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//</ol>
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//
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//<p> </p>
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//
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//<p><strong>示例 1：</strong></p>
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//
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//<p><img src="https://assets.leetcode.com/uploads/2021/07/24/lc-pickrandomrec.jpg" style="height: 539px; width: 419px;" /></p>
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//
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//<pre>
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//<strong>输入:
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//</strong>["Solution", "pick", "pick", "pick", "pick", "pick"]
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//[[[[-2, -2, 1, 1], [2, 2, 4, 6]]], [], [], [], [], []]
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//<strong>输出:
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//</strong>[null, [1, -2], [1, -1], [-1, -2], [-2, -2], [0, 0]]
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//
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//<strong>解释：</strong>
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//Solution solution = new Solution([[-2, -2, 1, 1], [2, 2, 4, 6]]);
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//solution.pick(); // 返回 [1, -2]
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//solution.pick(); // 返回 [1, -1]
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//solution.pick(); // 返回 [-1, -2]
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//solution.pick(); // 返回 [-2, -2]
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//solution.pick(); // 返回 [0, 0]</pre>
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//
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//<p> </p>
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//
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//<p><strong>提示：</strong></p>
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//
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//<ul>
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// <li><code>1 <= rects.length <= 100</code></li>
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// <li><code>rects[i].length == 4</code></li>
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// <li><code>-10<sup>9</sup> <= a<sub>i</sub> < x<sub>i</sub> <= 10<sup>9</sup></code></li>
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// <li><code>-10<sup>9</sup> <= b<sub>i</sub> < y<sub>i</sub> <= 10<sup>9</sup></code></li>
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// <li><code>x<sub>i</sub> - a<sub>i</sub> <= 2000</code></li>
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// <li><code>y<sub>i</sub> - b<sub>i</sub> <= 2000</code></li>
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// <li>所有的矩形不重叠。</li>
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// <li><code>pick</code> 最多被调用 <code>10<sup>4</sup></code> 次。</li>
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//</ul>
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//<div><div>Related Topics</div><div><li>水塘抽样</li><li>数学</li><li>二分查找</li><li>有序集合</li><li>前缀和</li><li>随机化</li></div></div><br><div><li>👍 98</li><li>👎 0</li></div>
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package leetcode.editor.cn;
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import java.util.Random;
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// 497:非重叠矩形中的随机点
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public class RandomPointInNonOverlappingRectangles {
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public static void main(String[] args) {
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// TO TEST
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}
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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int[][] rects;
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int[] mul;
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int count;
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Random random = new Random();
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public Solution(int[][] rects) {
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this.rects = rects;
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count = this.rects.length;
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mul = new int[count + 1];
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for (int i = 1; i <= count; i++) {
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mul[i] = mul[i - 1] + (this.rects[i - 1][2] - this.rects[i - 1][0] + 1) *
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(this.rects[i - 1][3] - this.rects[i - 1][1] + 1);
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}
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}
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public int[] pick() {
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int index = random.nextInt(mul[count]) + 1;
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int left = 0;
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int right = count;
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while (left < right) {
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int mid = left + right >> 1;
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if (mul[mid] >= index) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return new int[]{
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random.nextInt(rects[right - 1][2] - rects[right - 1][0] + 1) + rects[right - 1][0],
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random.nextInt(rects[right - 1][3] - rects[right - 1][1] + 1) + rects[right - 1][1]
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};
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}
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}
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/**
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* Your Solution object will be instantiated and called as such:
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* Solution obj = new Solution(rects);
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* int[] param_1 = obj.pick();
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*/
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//leetcode submit region end(Prohibit modification and deletion)
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}
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@ -0,0 +1,52 @@
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<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>
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<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>
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<p><strong>请注意 </strong>，整数点是具有整数坐标的点。</p>
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<p>实现 <code>Solution</code> 类:</p>
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<ul>
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<li><code>Solution(int[][] rects)</code> 用给定的矩形数组 <code>rects</code> 初始化对象。</li>
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<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>
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</ul>
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<ol>
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</ol>
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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://assets.leetcode.com/uploads/2021/07/24/lc-pickrandomrec.jpg" style="height: 539px; width: 419px;" /></p>
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<pre>
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<strong>输入:
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</strong>["Solution", "pick", "pick", "pick", "pick", "pick"]
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[[[[-2, -2, 1, 1], [2, 2, 4, 6]]], [], [], [], [], []]
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<strong>输出:
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</strong>[null, [1, -2], [1, -1], [-1, -2], [-2, -2], [0, 0]]
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<strong>解释：</strong>
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Solution solution = new Solution([[-2, -2, 1, 1], [2, 2, 4, 6]]);
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solution.pick(); // 返回 [1, -2]
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solution.pick(); // 返回 [1, -1]
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solution.pick(); // 返回 [-1, -2]
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solution.pick(); // 返回 [-2, -2]
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solution.pick(); // 返回 [0, 0]</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>1 <= rects.length <= 100</code></li>
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<li><code>rects[i].length == 4</code></li>
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<li><code>-10<sup>9</sup> <= a<sub>i</sub> < x<sub>i</sub> <= 10<sup>9</sup></code></li>
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<li><code>-10<sup>9</sup> <= b<sub>i</sub> < y<sub>i</sub> <= 10<sup>9</sup></code></li>
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<li><code>x<sub>i</sub> - a<sub>i</sub> <= 2000</code></li>
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<li><code>y<sub>i</sub> - b<sub>i</sub> <= 2000</code></li>
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<li>所有的矩形不重叠。</li>
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<li><code>pick</code> 最多被调用 <code>10<sup>4</sup></code> 次。</li>
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</ul>
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<div><div>Related Topics</div><div><li>水塘抽样</li><li>数学</li><li>二分查找</li><li>有序集合</li><li>前缀和</li><li>随机化</li></div></div><br><div><li>👍 98</li><li>👎 0</li></div>
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