From f702f0482d678ef2c989d477157b0b6542852348 Mon Sep 17 00:00:00 2001
From: bqlin <bqlins@163.com>
Date: Sat, 11 Dec 2021 18:18:45 +0800
Subject: [PATCH] =?UTF-8?q?0216.=E7=BB=84=E5=90=88=E6=80=BB=E5=92=8CIII?=
 =?UTF-8?q?=EF=BC=9A=E4=BC=98=E5=8C=96=E6=8E=92=E7=89=88=EF=BC=8C=E8=A1=A5?=
 =?UTF-8?q?=E5=85=85Swift=E7=89=88=E6=9C=AC?=
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---
 problems/0216.组合总和III.md | 45 +++++++++++++++++++++++++++-----
 1 file changed, 39 insertions(+), 6 deletions(-)

diff --git a/problems/0216.组合总和III.md b/problems/0216.组合总和III.md
index fa2ce37f..76d31098 100644
--- a/problems/0216.组合总和III.md
+++ b/problems/0216.组合总和III.md
@@ -57,7 +57,7 @@
 
 至于为什么取名为path？从上面树形结构中，可以看出，结果其实就是一条根节点到叶子节点的路径。
 
-```
+```cpp
 vector<vector<int>> result; // 存放结果集
 vector<int> path; // 符合条件的结果
 ```
@@ -71,7 +71,7 @@ vector<int> path; // 符合条件的结果
 
 所以代码如下：
 
-```
+```cpp
 vector<vector<int>> result;
 vector<int> path;
 void backtracking(int targetSum, int k, int sum, int startIndex)
@@ -168,7 +168,7 @@ public:
 
 那么剪枝的地方一定是在递归终止的地方剪，剪枝代码如下：
 
-```
+```cpp
 if (sum > targetSum) { // 剪枝操作
     return;
 }
@@ -319,7 +319,7 @@ class Solution:
         return res
 ```
 
-## Go：
+## Go
 
 回溯+减枝
 
@@ -351,7 +351,7 @@ func backTree(n,k,startIndex int,track *[]int,result *[][]int){
 }
 ```
 
-## javaScript:
+## javaScript
 
 ```js
 // 等差数列
@@ -390,7 +390,8 @@ var combinationSum3 = function(k, n) {
 };
 ```
 
-C:
+## C
+
 ```c
 int* path;
 int pathTop;
@@ -448,5 +449,37 @@ int** combinationSum3(int k, int n, int* returnSize, int** returnColumnSizes){
 }
 ```
 
+## Swift
+
+```swift
+func combinationSum3(_ k: Int, _ n: Int) -> [[Int]] {
+    var result = [[Int]]()
+    var path = [Int]()
+    func backtracking(targetSum: Int, k: Int, sum: Int, startIndex: Int) {
+        // 剪枝
+        if sum  > targetSum { return }
+        // 终止条件
+        if path.count == k {
+            if sum == targetSum {
+                result.append(path)
+            }
+            return
+        }
+
+        // 单层逻辑
+        let endIndex = 9
+        guard startIndex <= endIndex else { return }
+        for i in startIndex ... endIndex {
+            path.append(i) // 处理
+            backtracking(targetSum: targetSum, k: k, sum: sum + i, startIndex: i + 1)
+            path.removeLast() // 回溯
+        }
+    }
+
+    backtracking(targetSum: n, k: k, sum: 0, startIndex: 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>