添加 problem 497

Algorithms/0497. Random Point in Non-overlapping Rectangles/497. Random Point in Non-overlapping Rectangles.go

@@ -0,0 +1,63 @@
+package leetcode
+
+import "math/rand"
+
+// Solution497 define
+type Solution497 struct {
+	rects [][]int
+	arr   []int
+}
+
+// Constructor497 define
+func Constructor497(rects [][]int) Solution497 {
+	s := Solution497{
+		rects: rects,
+		arr:   make([]int, len(rects)),
+	}
+
+	for i := 0; i < len(rects); i++ {
+		area := (rects[i][2] - rects[i][0] + 1) * (rects[i][3] - rects[i][1] + 1)
+		if area < 0 {
+			area = -area
+		}
+		if i == 0 {
+			s.arr[0] = area
+		} else {
+			s.arr[i] = s.arr[i-1] + area
+		}
+	}
+	return s
+}
+
+// Pick define
+func (so *Solution497) Pick() []int {
+	r := rand.Int() % so.arr[len(so.arr)-1]
+	//get rectangle first
+	low, high, index := 0, len(so.arr)-1, -1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if so.arr[mid] > r {
+			if mid == 0 || so.arr[mid-1] <= r {
+				index = mid
+				break
+			}
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	if index == -1 {
+		index = low
+	}
+	if index > 0 {
+		r = r - so.arr[index-1]
+	}
+	length := so.rects[index][2] - so.rects[index][0]
+	return []int{so.rects[index][0] + r%(length+1), so.rects[index][1] + r/(length+1)}
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(rects);
+ * param_1 := obj.Pick();
+ */

Algorithms/0497. Random Point in Non-overlapping Rectangles/497. Random Point in Non-overlapping Rectangles_test.go

@@ -0,0 +1,17 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem497(t *testing.T) {
+	w := [][]int{[]int{1, 1, 5, 5}}
+	sol := Constructor497(w)
+	fmt.Printf("1.Pick = %v\n", sol.Pick())
+	fmt.Printf("2.Pick = %v\n", sol.Pick())
+	fmt.Printf("3.Pick = %v\n", sol.Pick())
+	fmt.Printf("4.Pick = %v\n", sol.Pick())
+	fmt.Printf("5.Pick = %v\n", sol.Pick())
+	fmt.Printf("6.Pick = %v\n", sol.Pick())
+}

Algorithms/0497. Random Point in Non-overlapping Rectangles/README.md

@@ -0,0 +1,63 @@
+# [497. Random Point in Non-overlapping Rectangles](https://leetcode.com/problems/random-point-in-non-overlapping-rectangles)
+
+
+## 题目:
+
+Given a list of **non-overlapping** axis-aligned rectangles `rects`, write a function `pick` which randomly and uniformily picks an **integer point** in the space covered by the rectangles.
+
+Note:
+
+1. An **integer point** is a point that has integer coordinates.
+2. A point on the perimeter of a rectangle is **included** in the space covered by the rectangles.
+3. `i`th rectangle = `rects[i]` = `[x1,y1,x2,y2]`, where `[x1, y1]` are the integer coordinates of the bottom-left corner, and `[x2, y2]` are the integer coordinates of the top-right corner.
+4. length and width of each rectangle does not exceed `2000`.
+5. `1 <= rects.length <= 100`
+6. `pick` return a point as an array of integer coordinates `[p_x, p_y]`
+7. `pick` is called at most `10000` times.
+
+**Example 1:**
+
+    Input: 
+    ["Solution","pick","pick","pick"]
+    [[[[1,1,5,5]]],[],[],[]]
+    Output: 
+    [null,[4,1],[4,1],[3,3]]
+
+**Example 2:**
+
+    Input: 
+    ["Solution","pick","pick","pick","pick","pick"]
+    [[[[-2,-2,-1,-1],[1,0,3,0]]],[],[],[],[],[]]
+    Output: 
+    [null,[-1,-2],[2,0],[-2,-1],[3,0],[-2,-2]]
+
+**Explanation of Input Syntax:**
+
+The input is two lists: the subroutines called and their arguments. `Solution`'s constructor has one argument, the array of rectangles `rects`. `pick` has no arguments. Arguments are always wrapped with a list, even if there aren't any.
+
+
+## 题目大意
+
+给定一个非重叠轴对齐矩形的列表 rects，写一个函数 pick 随机均匀地选取矩形覆盖的空间中的整数点。
+
+提示：
+
+1. 整数点是具有整数坐标的点。
+2. 矩形周边上的点包含在矩形覆盖的空间中。
+3. 第 i 个矩形 rects [i] = [x1，y1，x2，y2]，其中 [x1，y1] 是左下角的整数坐标，[x2，y2] 是右上角的整数坐标。
+4. 每个矩形的长度和宽度不超过 2000。
+5. 1 <= rects.length <= 100
+6. pick 以整数坐标数组 [p_x, p_y] 的形式返回一个点。
+7. pick 最多被调用10000次。
+
+
+输入语法的说明：
+
+输入是两个列表：调用的子例程及其参数。Solution 的构造函数有一个参数，即矩形数组 rects。pick 没有参数。参数总是用列表包装的，即使没有也是如此。
+
+
+## 解题思路
+
+
+- 给出一个非重叠轴对齐矩形列表，每个矩形用左下角和右上角的两个坐标表示。要求 `pick()` 随机均匀地选取矩形覆盖的空间中的整数点。
+- 这一题是第 528 题的变种题，这一题权重是面积，按权重（面积）选择一个矩形，然后再从矩形中随机选择一个点即可。思路和代码和第 528 题一样。