From c99bf39601a06278f9cca2e5f6d61638e23f3ced Mon Sep 17 00:00:00 2001
From: ZongqinWang <1722249371@qq.com>
Date: Thu, 23 Jun 2022 21:34:24 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200474.=E4=B8=80=E5=92=8C?=
 =?UTF-8?q?=E9=9B=B6.md=20Scala=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0474.一和零.md | 83 +++++++++++++++++++++++++++++++++++---
 1 file changed, 78 insertions(+), 5 deletions(-)

diff --git a/problems/0474.一和零.md b/problems/0474.一和零.md
index d38ce03f..d6c598aa 100644
--- a/problems/0474.一和零.md
+++ b/problems/0474.一和零.md
@@ -163,7 +163,7 @@ public:
 ## 其他语言版本
 
 
-Java：
+### Java
 ```Java
 class Solution {
     public int findMaxForm(String[] strs, int m, int n) {
@@ -192,7 +192,7 @@ class Solution {
 }
 ```
 
-Python：
+### Python
 ```python
 class Solution:
     def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
@@ -208,7 +208,7 @@ class Solution:
         return dp[m][n]
 ```
 
-Go：
+### Go
 ```go
 func findMaxForm(strs []string, m int, n int) int {
 	// 定义数组
@@ -294,7 +294,7 @@ func getMax(a,b int)int{
 }
 ```
 
-Javascript：
+### Javascript
 ```javascript
 const findMaxForm = (strs, m, n) => {
     const dp = Array.from(Array(m+1), () => Array(n+1).fill(0));
@@ -323,7 +323,7 @@ const findMaxForm = (strs, m, n) => {
 };
 ```
 
-TypeScript：
+### TypeScript
 
 > 滚动数组，二维数组法
 
@@ -446,7 +446,80 @@ function isValidSubSet(strs: string[], m: number, n: number): boolean {
 }
 ```
 
+### Scala
 
+背包：
+```scala
+object Solution {
+  def findMaxForm(strs: Array[String], m: Int, n: Int): Int = {
+    var dp = Array.ofDim[Int](m + 1, n + 1)
+
+    var (oneNum, zeroNum) = (0, 0)
+
+    for (str <- strs) {
+      oneNum = 0
+      zeroNum = 0
+      for (i <- str.indices) {
+        if (str(i) == '0') zeroNum += 1
+        else oneNum += 1
+      }
+
+      for (i <- m to zeroNum by -1) {
+        for (j <- n to oneNum by -1) {
+          dp(i)(j) = math.max(dp(i)(j), dp(i - zeroNum)(j - oneNum) + 1)
+        }
+      }
+    }
+
+    dp(m)(n)
+  }
+}
+```
+
+回溯法(超时)：
+```scala
+object Solution {
+  import scala.collection.mutable
+
+  var res = Int.MinValue
+
+  def test(str: String): (Int, Int) = {
+    var (zero, one) = (0, 0)
+    for (i <- str.indices) {
+      if (str(i) == '1') one += 1
+      else zero += 1
+    }
+    (zero, one)
+  }
+
+  def travsel(strs: Array[String], path: mutable.ArrayBuffer[String], m: Int, n: Int, startIndex: Int): Unit = {
+    if (startIndex > strs.length) {
+      return
+    }
+
+    res = math.max(res, path.length)
+
+    for (i <- startIndex until strs.length) {
+
+      var (zero, one) = test(strs(i))
+
+      // 如果0的个数小于m，1的个数小于n，则可以回溯
+      if (zero <= m && one <= n) {
+        path.append(strs(i))
+        travsel(strs, path, m - zero, n - one, i + 1)
+        path.remove(path.length - 1)
+      }
+    }
+  }
+
+  def findMaxForm(strs: Array[String], m: Int, n: Int): Int = {
+    res = Int.MinValue
+    var path = mutable.ArrayBuffer[String]()
+    travsel(strs, path, m, n, 0)
+    res
+  }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>