Add solution 478

leetcode/0115.Distinct-Subsequences/README.md

@@ -44,9 +44,7 @@ babgbagbabgbagbabgbagbabgbagbabgbag
 - 当 `i < len(s)` 且 `j < len(t)` 的时候，如果 `s[i] == t[j]`，有 2 种匹配方式，第一种将 `s[i]` 与 `t[j]` 匹配，那么 `t[j+1:]` 匹配 `s[i+1:]` 的子序列，子序列数为 `dp[i+1][j+1]`；第二种将 `s[i]` 不与 `t[j]` 匹配，`t[j:]` 作为 `s[i+1:]` 的子序列，子序列数为 `dp[i+1][j]`。综合 2 种情况，当 `s[i] == t[j]` 时，`dp[i][j] = dp[i+1][j+1] + dp[i+1][j]`。
 - 如果 `s[i] != t[j]`，此时 `t[j:]` 只能作为 `s[i+1:]` 的子序列，子序列数为 `dp[i+1][j]`。所以当 `s[i] != t[j]` 时，`dp[i][j] = dp[i+1][j]`。综上分析得：
 
-	{{< katex display >}}
-    dp[i][j] = \left\{\begin{matrix}dp[i+1][j+1]+dp[i+1][j]&,s[i]=t[j]\\ dp[i+1][j]&,s[i]!=t[j]\end{matrix}\right.
-	{{< /katex >}}
+	$$dp[i][j] = \left\{\begin{matrix}dp[i+1][j+1]+dp[i+1][j]&,s[i]=t[j]\\ dp[i+1][j]&,s[i]!=t[j]\end{matrix}\right.$$
 
 - 最后是优化版本。写出上述代码以后，可以发现填表的过程是从右下角一直填到左上角。填表顺序是 从下往上一行一行的填。行内从右往左填。于是可以将这个二维数据压缩到一维。因为填充当前行只需要用到它的下一行信息即可，更进一步，用到的是下一行中右边元素的信息。于是可以每次更新该行时，先将旧的值存起来，计算更新该行的时候从右往左更新。这样做即可减少一维空间，将原来的二维数组压缩到一维数组。
 

leetcode/0478.Generate-Random-Point-in-a-Circle/478. Generate Random Point in a Circle.go

@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"math"
+	"math/rand"
+	"time"
+)
+
+type Solution struct {
+	r float64
+	x float64
+	y float64
+}
+
+func Constructor(radius float64, x_center float64, y_center float64) Solution {
+	rand.Seed(time.Now().UnixNano())
+	return Solution{radius, x_center, y_center}
+}
+
+func (this *Solution) RandPoint() []float64 {
+	/*
+	   a := angle()
+	   r := this.r * math.Sqrt(rand.Float64())
+	   x := r * math.Cos(a) + this.x
+	   y := r * math.Sin(a) + this.y
+	   return []float64{x, y}*/
+	for {
+		rx := 2*rand.Float64() - 1.0
+		ry := 2*rand.Float64() - 1.0
+		x := this.r * rx
+		y := this.r * ry
+		if x*x+y*y <= this.r*this.r {
+			return []float64{x + this.x, y + this.y}
+		}
+	}
+}
+
+func angle() float64 {
+	return rand.Float64() * 2 * math.Pi
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(radius, x_center, y_center);
+ * param_1 := obj.RandPoint();
+ */

leetcode/0478.Generate-Random-Point-in-a-Circle/478. Generate Random Point in a Circle_test.go

@@ -0,0 +1,18 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+func Test_Problem478(t *testing.T) {
+	obj := Constructor(1, 0, 0)
+	fmt.Printf("RandPoint() = %v\n", obj.RandPoint())
+	fmt.Printf("RandPoint() = %v\n", obj.RandPoint())
+	fmt.Printf("RandPoint() = %v\n", obj.RandPoint())
+
+	obj = Constructor(10, 5, -7.5)
+	fmt.Printf("RandPoint() = %v\n", obj.RandPoint())
+	fmt.Printf("RandPoint() = %v\n", obj.RandPoint())
+	fmt.Printf("RandPoint() = %v\n", obj.RandPoint())
+}

leetcode/0478.Generate-Random-Point-in-a-Circle/README.md

@@ -0,0 +1,103 @@
+# [478. Generate Random Point in a Circle](https://leetcode.com/problems/generate-random-point-in-a-circle/)
+
+
+## 题目
+
+Given the radius and x-y positions of the center of a circle, write a function `randPoint` which generates a uniform random point in the circle.
+
+Note:
+
+1. input and output values are in [floating-point](https://www.webopedia.com/TERM/F/floating_point_number.html).
+2. radius and x-y position of the center of the circle is passed into the class constructor.
+3. a point on the circumference of the circle is considered to be in the circle.
+4. `randPoint` returns a size 2 array containing x-position and y-position of the random point, in that order.
+
+**Example 1:**
+
+```
+Input: 
+["Solution","randPoint","randPoint","randPoint"]
+[[1,0,0],[],[],[]]
+Output: [null,[-0.72939,-0.65505],[-0.78502,-0.28626],[-0.83119,-0.19803]]
+
+```
+
+**Example 2:**
+
+```
+Input: 
+["Solution","randPoint","randPoint","randPoint"]
+[[10,5,-7.5],[],[],[]]
+Output: [null,[11.52438,-8.33273],[2.46992,-16.21705],[11.13430,-12.42337]]
+```
+
+**Explanation of Input Syntax:**
+
+The input is two lists: the subroutines called and their arguments. `Solution`'s constructor has three arguments, the radius, x-position of the center, and y-position of the center of the circle. `randPoint` has no arguments. Arguments are always wrapped with a list, even if there aren't any.
+
+## 题目大意
+
+给定圆的半径和圆心的 x、y 坐标，写一个在圆中产生均匀随机点的函数 randPoint 。
+
+说明:
+
+- 输入值和输出值都将是浮点数。
+- 圆的半径和圆心的 x、y 坐标将作为参数传递给类的构造函数。
+- 圆周上的点也认为是在圆中。
+- randPoint 返回一个包含随机点的x坐标和y坐标的大小为2的数组。
+
+## 解题思路
+
+- 随机产生一个圆内的点，这个点一定满足定义 `(x-a)^2+(y-b)^2 ≤ R^2`，其中 `(a,b)` 是圆的圆心坐标，`R` 是半径。
+- 先假设圆心坐标在 (0,0)，这样方便计算，最终输出坐标的时候整体加上圆心的偏移量即可。`rand.Float64()` 产生一个 `[0.0,1.0)` 区间的浮点数。`-R ≤ 2 * R * rand() - R < R`，利用随机产生坐标点的横纵坐标 `(x,y)` 与半径 R 的关系，如果 `x^2 + y^2 ≤ R^2`，那么说明产生的点在圆内。最终输出的时候要记得加上圆心坐标的偏移值。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math"
+	"math/rand"
+	"time"
+)
+
+type Solution struct {
+	r float64
+	x float64
+	y float64
+}
+
+func Constructor(radius float64, x_center float64, y_center float64) Solution {
+	rand.Seed(time.Now().UnixNano())
+	return Solution{radius, x_center, y_center}
+}
+
+func (this *Solution) RandPoint() []float64 {
+	/*
+	   a := angle()
+	   r := this.r * math.Sqrt(rand.Float64())
+	   x := r * math.Cos(a) + this.x
+	   y := r * math.Sin(a) + this.y
+	   return []float64{x, y}*/
+	for {
+		rx := 2*rand.Float64() - 1.0
+		ry := 2*rand.Float64() - 1.0
+		x := this.r * rx
+		y := this.r * ry
+		if x*x+y*y <= this.r*this.r {
+			return []float64{x + this.x, y + this.y}
+		}
+	}
+}
+
+func angle() float64 {
+	return rand.Float64() * 2 * math.Pi
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(radius, x_center, y_center);
+ * param_1 := obj.RandPoint();
+ */
+```