From a1359e0f3b2578197eff747441d816aaa9ae2a7b Mon Sep 17 00:00:00 2001
From: bqlin <bqlins@163.com>
Date: Sat, 11 Dec 2021 18:23:33 +0800
Subject: [PATCH] =?UTF-8?q?0077.=E7=BB=84=E5=90=88=EF=BC=9A=E4=BC=98?=
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---
 problems/0077.组合.md | 42 +++++++++++++++++++++++++++++++++++------
 1 file changed, 36 insertions(+), 6 deletions(-)

diff --git a/problems/0077.组合.md b/problems/0077.组合.md
index 16313cb8..6c74ab52 100644
--- a/problems/0077.组合.md
+++ b/problems/0077.组合.md
@@ -109,7 +109,7 @@ for (int i = 1; i <= n; i++) {
 
 代码如下：
 
-```
+```cpp
 vector<vector<int>> result; // 存放符合条件结果的集合
 vector<int> path; // 用来存放符合条件结果
 ```
@@ -132,7 +132,7 @@ vector<int> path; // 用来存放符合条件结果
 
 那么整体代码如下：
 
-```
+```cpp
 vector<vector<int>> result; // 存放符合条件结果的集合
 vector<int> path; // 用来存放符合条件单一结果
 void backtracking(int n, int k, int startIndex) 
@@ -152,7 +152,7 @@ path这个数组的大小如果达到k，说明我们找到了一个子集大小
 
 所以终止条件代码如下：
 
-```
+```cpp
 if (path.size() == k) {
     result.push_back(path);
     return;
@@ -248,7 +248,7 @@ void backtracking(参数) {
 
 在遍历的过程中有如下代码：
 
-```
+```cpp
 for (int i = startIndex; i <= n; i++) {
     path.push_back(i);
     backtracking(n, k, i + 1);
@@ -411,7 +411,7 @@ class Solution:
                 path.pop()
         backtrack(n, k, 1)
         return res
-``` 
+```
 
 剪枝：
 ```python3
@@ -454,7 +454,7 @@ const combineHelper = (n, k, startIndex) => {
     path.pop()
   }
 }
-``` 
+```
 
 
 
@@ -621,5 +621,35 @@ int** combine(int n, int k, int* returnSize, int** returnColumnSizes){
 }
 ```
 
+## Swift
+
+```swift
+func combine(_ n: Int, _ k: Int) -> [[Int]] {
+    var path = [Int]()
+    var result = [[Int]]()
+    func backtracking(_ n: Int, _ k: Int, _ startIndex: Int) {
+        // 结束条件，并收集结果
+        if path.count == k {
+            result.append(path)
+            return
+        }
+
+        // 单层逻辑
+        // let end = n
+        // 剪枝优化
+        let end = n - (k - path.count) + 1
+        guard startIndex <= end else { return }
+        for i in startIndex ... end {
+            path.append(i) // 处理结点
+            backtracking(n, k, i + 1) // 递归
+            path.removeLast() // 回溯
+        }
+    }
+
+    backtracking(n, k, 1)
+    return result
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>