From b884383b1264da0bf158bcf34ffba62afa754c36 Mon Sep 17 00:00:00 2001
From: bqlin <bqlins@163.com>
Date: Sun, 12 Dec 2021 18:35:29 +0800
Subject: [PATCH] =?UTF-8?q?0090.=E5=AD=90=E9=9B=86II=EF=BC=9A=E4=BC=98?=
 =?UTF-8?q?=E5=8C=96=E6=8E=92=E7=89=88=EF=BC=8C=E8=A1=A5=E5=85=85Swift?=
 =?UTF-8?q?=E7=89=88=E6=9C=AC?=
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---
 problems/0090.子集II.md | 31 +++++++++++++++++++++++++++----
 1 file changed, 27 insertions(+), 4 deletions(-)

diff --git a/problems/0090.子集II.md b/problems/0090.子集II.md
index 95b6080b..17612801 100644
--- a/problems/0090.子集II.md
+++ b/problems/0090.子集II.md
@@ -156,14 +156,12 @@ public:
 
 当然本题去重的逻辑，也可以这么写
 
-```
+```cpp
 if (i > startIndex && nums[i] == nums[i - 1] ) {
-        continue;
+	continue;
 }
 ```
 
-
-
 ## 其他语言版本
 
 
@@ -359,6 +357,31 @@ int** subsetsWithDup(int* nums, int numsSize, int* returnSize, int** returnColum
 }
 ```
 
+## Swift
+
+```swift
+func subsetsWithDup(_ nums: [Int]) -> [[Int]] {
+    let nums = nums.sorted()
+    var result = [[Int]]()
+    var path = [Int]()
+    func backtracking(startIndex: Int) {
+        // 直接收集结果
+        result.append(path)
+
+        let end = nums.count
+        guard startIndex < end else { return } // 终止条件
+        for i in startIndex ..< end {
+            if i > startIndex, nums[i] == nums[i - 1] { continue } // 跳过重复元素
+            path.append(nums[i]) // 处理：收集元素
+            backtracking(startIndex: i + 1) // 元素不重复访问
+            path.removeLast() // 回溯
+        }
+    }
+    backtracking(startIndex: 0)
+    return result
+}
+```
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>