From 97fc88e533418bf9070bd9fb549a23a2499805b4 Mon Sep 17 00:00:00 2001
From: ZongqinWang <1722249371@qq.com>
Date: Tue, 7 Jun 2022 16:59:55 +0800
Subject: [PATCH 1/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200046.=E5=85=A8?=
=?UTF-8?q?=E6=8E=92=E5=88=97.md=20Scala=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0046.全排列.md | 30 ++++++++++++++++++++++++++++++
1 file changed, 30 insertions(+)
diff --git a/problems/0046.全排列.md b/problems/0046.全排列.md
index 836c3646..ec3adaa7 100644
--- a/problems/0046.全排列.md
+++ b/problems/0046.全排列.md
@@ -456,6 +456,36 @@ func permute(_ nums: [Int]) -> [[Int]] {
}
```
+### Scala
+
+```scala
+object Solution {
+ import scala.collection.mutable
+ def permute(nums: Array[Int]): List[List[Int]] = {
+ var result = mutable.ListBuffer[List[Int]]()
+ var path = mutable.ListBuffer[Int]()
+
+ def backtracking(used: Array[Boolean]): Unit = {
+ if (path.size == nums.size) {
+ // 如果path的长度和nums相等，那么可以添加到结果集
+ result.append(path.toList)
+ return
+ }
+ // 添加循环守卫，只有当当前数字没有用过的情况下才进入回溯
+ for (i <- nums.indices if used(i) == false) {
+ used(i) = true
+ path.append(nums(i))
+ backtracking(used) // 回溯
+ path.remove(path.size - 1)
+ used(i) = false
+ }
+ }
+
+ backtracking(new Array[Boolean](nums.size)) // 调用方法
+ result.toList // 最终返回结果集的List形式
+ }
+}
+```
-----------------------
From 8537401616f10ce85038bbd94aa951a20bb877d6 Mon Sep 17 00:00:00 2001
From: ZongqinWang <1722249371@qq.com>
Date: Tue, 7 Jun 2022 17:24:41 +0800
Subject: [PATCH 2/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200047.=E5=85=A8?=
=?UTF-8?q?=E6=8E=92=E5=88=97II.md=20Scala=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0047.全排列II.md | 38 ++++++++++++++++++++++++++++++++++++
1 file changed, 38 insertions(+)
diff --git a/problems/0047.全排列II.md b/problems/0047.全排列II.md
index cce25cd9..0a5debcc 100644
--- a/problems/0047.全排列II.md
+++ b/problems/0047.全排列II.md
@@ -422,5 +422,43 @@ int** permuteUnique(int* nums, int numsSize, int* returnSize, int** returnColumn
}
```
+### Scala
+
+```scala
+object Solution {
+ import scala.collection.mutable
+ def permuteUnique(nums: Array[Int]): List[List[Int]] = {
+ var result = mutable.ListBuffer[List[Int]]()
+ var path = mutable.ListBuffer[Int]()
+ var num = nums.sorted // 首先对数据进行排序
+
+ def backtracking(used: Array[Boolean]): Unit = {
+ if (path.size == num.size) {
+ // 如果path的size等于num了，那么可以添加到结果集
+ result.append(path.toList)
+ return
+ }
+ // 循环守卫，当前元素没被使用过就进入循环体
+ for (i <- num.indices if used(i) == false) {
+ // 当前索引为0，不存在和前一个数字相等可以进入回溯
+ // 当前索引值和上一个索引不相等，可以回溯
+ // 前一个索引对应的值没有被选，可以回溯
+ // 因为Scala没有continue，只能将逻辑反过来写
+ if (i == 0 || (i > 0 && num(i) != num(i - 1)) || used(i-1) == false) {
+ used(i) = true
+ path.append(num(i))
+ backtracking(used)
+ path.remove(path.size - 1)
+ used(i) = false
+ }
+ }
+ }
+
+ backtracking(new Array[Boolean](nums.length))
+ result.toList
+ }
+}
+```
+
-----------------------
