add scala 96 *

scala-leetcode/README.md

@@ -48,6 +48,7 @@ Leetcode Scala 实现
 * [BiNode @unchecked](./src/main/scala/io/github/dreamylost/Leetcode_Interview_17_12.scala)
 * [二叉树的中序遍历 非递归](./src/main/scala/io/github/dreamylost/Leetcode_94.scala)
 * [不同的二叉搜索树 II](./src/main/scala/io/github/dreamylost/Leetcode_95.scala)
+* [不同的二叉搜索树](./src/main/scala/io/github/dreamylost/Leetcode_96.scala)
 
 ## 链表
 

scala-leetcode/src/main/scala/io/github/dreamylost/Leetcode_96.scala

+package io.github.dreamylost
+
+import scala.collection.immutable.HashMap
+
+/**
+  * 96. 不同的二叉搜索树
+  *
+  * 给定一个整数 n，求以 1 ... n 为节点组成的二叉搜索树有多少种？
+  *
+  * @author 梦境迷离 dreamylost
+  * @since 2020-06-28
+  * @version v1.0
+  */
+object Leetcode_96 extends App {
+
+  val ret = numTrees(3)
+  val ret2 = numTrees2(3)
+  val ret3 = numTrees3(3)
+  val ret4 = numTrees4(3)
+  println(ret)
+  println(ret2)
+  println(ret3)
+  println(ret4)
+
+  /**
+    * 不能直接使用95的代码，会超时
+    * 利用递推公式求解
+    *
+    * fn=f(1)+f(2)+f(3)+...fn(n)
+    * 对任意的i，i的左子树有i-1个节点,i的右子树有n-i个节点
+    * f(i)=f(i-1)*f(n-i)
+    * 代入到fn中，得f(n) = f(0)*f(n-1)+f(1)f(n-2)+f(2)f(n-3)+...+f(n-1)f(0)
+    *
+    * 4600 ms,25.00%(直接用公式太慢了)
+    * 49.2 MB,100.00%
+    *
+    * @param n
+    * @return
+    */
+  def numTrees(n: Int): Int = {
+    //根据地推公式直接递归
+    var ret = 0
+    if (n == 1 || n == 0) 1
+    else {
+      for (i <- 1 to n) {
+        ret += numTrees(i - 1) * numTrees(n - i)
+      }
+      ret
+    }
+  }
+
+  /**
+    * 动态规划
+    * 460 ms,50.00%
+    * 49 MB,100.00%
+    *
+    * @param n
+    * @return
+    */
+  def numTrees2(n: Int): Int = {
+    if (n == 1 || n == 0) return 1
+    val ret = new Array[Int](n + 1)
+    ret(0) = 1
+    for (i <- 1 to n) {
+      for (j <- 1 to i) {
+        ret.update(i, ret(j - 1) * ret(i - j) + ret(i))
+      }
+    }
+    ret(n)
+  }
+
+  /**
+    * 与95类似，二叉搜索树的种类只与个数有关，缓存个数
+    * 472 ms,50.00%
+    * 49.1 MB,100.00%
+    *
+    * @param n
+    * @return
+    */
+  def numTrees3(n: Int): Int = {
+    var map = HashMap.empty[Int, Int]
+
+    def buildSearchTree(left: Int, right: Int): Int = {
+      var sum = 0
+      if (map.contains(right - left)) {
+        return map(right - left)
+      }
+      if (left > right) {
+        1
+      } else {
+        (left to right).foreach { i =>
+          val lCount = buildSearchTree(left, i - 1)
+          val rCount = buildSearchTree(i + 1, right)
+          sum += lCount * rCount
+        }
+        map = map ++ Map(right - left -> sum)
+        sum
+      }
+    }
+
+    if (n < 1) 0 else buildSearchTree(1, n)
+  }
+
+  /**
+    * 上面的递推公式 = 卡特兰数 https://baike.baidu.com/item/catalan/7605685?fr=aladdin
+    *
+    * 468 ms,50.00%
+    * 48.8 MB,100.00%
+    *
+    * @param n
+    * @return
+    */
+  def numTrees4(n: Int): Int = {
+    var ret = 1L
+    for (i <- 0 until n) {
+      ret = ret * 2 * (2 * i + 1) / (i + 2)
+    }
+    ret.toInt
+  }
+}