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] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200047.=E5=85=A8=E6=8E=92?= =?UTF-8?q?=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 + } +} +``` + -----------------------