From 7e12d862ae6a4a0e7006d4b5968a7df510b42bd6 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E8=BD=A9=E8=BE=95=E9=BE=99=E5=84=BF?= Date: Thu, 9 Jun 2022 17:03:04 +0800 Subject: [PATCH] =?UTF-8?q?497:=E9=9D=9E=E9=87=8D=E5=8F=A0=E7=9F=A9?= =?UTF-8?q?=E5=BD=A2=E4=B8=AD=E7=9A=84=E9=9A=8F=E6=9C=BA=E7=82=B9?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...RandomPointInNonOverlappingRectangles.java | 106 ++++++++++++++++++ .../RandomPointInNonOverlappingRectangles.md | 52 +++++++++ 2 files changed, 158 insertions(+) create mode 100644 src/main/java/leetcode/editor/cn/RandomPointInNonOverlappingRectangles.java create mode 100644 src/main/java/leetcode/editor/cn/doc/content/RandomPointInNonOverlappingRectangles.md diff --git a/src/main/java/leetcode/editor/cn/RandomPointInNonOverlappingRectangles.java b/src/main/java/leetcode/editor/cn/RandomPointInNonOverlappingRectangles.java new file mode 100644 index 0000000..258763e --- /dev/null +++ b/src/main/java/leetcode/editor/cn/RandomPointInNonOverlappingRectangles.java @@ -0,0 +1,106 @@ +//

给定一个由非重叠的轴对齐矩形的数组 rects ，其中 rects[i] = [ai, bi, xi, yi] 表示 (ai, bi) 是第 i 个矩形的左下角点，(xi, yi) 是第 i 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。

+// +//

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。

+// +//

请注意 ，整数点是具有整数坐标的点。

+// +//

实现 Solution 类:

+// +// +// +//
    +//
+// +//

 

+// +//

示例 1：

+// +//

+// +//
+//输入: 
+//["Solution", "pick", "pick", "pick", "pick", "pick"]
+//[[[[-2, -2, 1, 1], [2, 2, 4, 6]]], [], [], [], [], []]
+//输出: 
+//[null, [1, -2], [1, -1], [-1, -2], [-2, -2], [0, 0]]
+//
+//解释：
+//Solution solution = new Solution([[-2, -2, 1, 1], [2, 2, 4, 6]]);
+//solution.pick(); // 返回 [1, -2]
+//solution.pick(); // 返回 [1, -1]
+//solution.pick(); // 返回 [-1, -2]
+//solution.pick(); // 返回 [-2, -2]
+//solution.pick(); // 返回 [0, 0]
+// +//

 

+// +//

提示：

+// +// +//
Related Topics
  • 水塘抽样
  • 数学
  • 二分查找
  • 有序集合
  • 前缀和
  • 随机化

  • 👍 98
  • 👎 0
    • +package leetcode.editor.cn; + +import java.util.Random; + +// 497:非重叠矩形中的随机点 +public class RandomPointInNonOverlappingRectangles { + public static void main(String[] args) { + // TO TEST + } + + //leetcode submit region begin(Prohibit modification and deletion) + class Solution { + int[][] rects; + int[] mul; + int count; + Random random = new Random(); + + public Solution(int[][] rects) { + this.rects = rects; + count = this.rects.length; + mul = new int[count + 1]; + for (int i = 1; i <= count; i++) { + mul[i] = mul[i - 1] + (this.rects[i - 1][2] - this.rects[i - 1][0] + 1) * + (this.rects[i - 1][3] - this.rects[i - 1][1] + 1); + } + } + + public int[] pick() { + int index = random.nextInt(mul[count]) + 1; + int left = 0; + int right = count; + while (left < right) { + int mid = left + right >> 1; + if (mul[mid] >= index) { + right = mid; + } else { + left = mid + 1; + } + } + return new int[]{ + random.nextInt(rects[right - 1][2] - rects[right - 1][0] + 1) + rects[right - 1][0], + random.nextInt(rects[right - 1][3] - rects[right - 1][1] + 1) + rects[right - 1][1] + }; + } + } + +/** + * Your Solution object will be instantiated and called as such: + * Solution obj = new Solution(rects); + * int[] param_1 = obj.pick(); + */ +//leetcode submit region end(Prohibit modification and deletion) + +} diff --git a/src/main/java/leetcode/editor/cn/doc/content/RandomPointInNonOverlappingRectangles.md b/src/main/java/leetcode/editor/cn/doc/content/RandomPointInNonOverlappingRectangles.md new file mode 100644 index 0000000..fb395fd --- /dev/null +++ b/src/main/java/leetcode/editor/cn/doc/content/RandomPointInNonOverlappingRectangles.md @@ -0,0 +1,52 @@ +

    给定一个由非重叠的轴对齐矩形的数组 rects ，其中 rects[i] = [ai, bi, xi, yi] 表示 (ai, bi) 是第 i 个矩形的左下角点，(xi, yi) 是第 i 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。

    + +

    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。

    + +

    请注意 ，整数点是具有整数坐标的点。

    + +

    实现 Solution 类:

    + + + +
      +
    + +

     

    + +

    示例 1：

    + +

    + +
    +输入: 
+["Solution", "pick", "pick", "pick", "pick", "pick"]
+[[[[-2, -2, 1, 1], [2, 2, 4, 6]]], [], [], [], [], []]
+输出: 
+[null, [1, -2], [1, -1], [-1, -2], [-2, -2], [0, 0]]
+
+解释：
+Solution solution = new Solution([[-2, -2, 1, 1], [2, 2, 4, 6]]);
+solution.pick(); // 返回 [1, -2]
+solution.pick(); // 返回 [1, -1]
+solution.pick(); // 返回 [-1, -2]
+solution.pick(); // 返回 [-2, -2]
+solution.pick(); // 返回 [0, 0]
    + +

     

    + +

    提示：

    + + +
    Related Topics
  • 水塘抽样
  • 数学
  • 二分查找
  • 有序集合
  • 前缀和
  • 随机化

  • 👍 98
  • 👎 0
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