From 77aeb31bad82702662be54ab77b7c640b6e30ca5 Mon Sep 17 00:00:00 2001
From: Yuhao Ju <juguaguaji@gmail.com>
Date: Wed, 7 Dec 2022 21:07:21 +0800
Subject: [PATCH] =?UTF-8?q?update=200216.=E7=BB=84=E5=90=88=E6=80=BB?=
 =?UTF-8?q?=E5=92=8CIII:=20=E6=9B=BF=E6=8D=A2=20go=20=E4=BB=A3=E7=A0=81?=
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---
 problems/0216.组合总和III.md | 41 +++++++++++++++++---------------
 1 file changed, 22 insertions(+), 19 deletions(-)

diff --git a/problems/0216.组合总和III.md b/problems/0216.组合总和III.md
index 9fc949cc..19fe1eb9 100644
--- a/problems/0216.组合总和III.md
+++ b/problems/0216.组合总和III.md
@@ -36,7 +36,7 @@
 
 想到这一点了，做过[77. 组合](https://programmercarl.com/0077.组合.html)之后，本题是简单一些了。
 
-本题k相当于了树的深度，9（因为整个集合就是9个数）就是树的宽度。
+本题k相当于树的深度，9（因为整个集合就是9个数）就是树的宽度。
 
 例如 k = 2，n = 4的话，就是在集合[1,2,3,4,5,6,7,8,9]中求 k（个数） = 2, n（和） = 4的组合。
 
@@ -380,29 +380,32 @@ class Solution:
 回溯+减枝
 
 ```go
+var (
+    res [][]int
+    path  []int
+)
 func combinationSum3(k int, n int) [][]int {
-    var track []int// 遍历路径
-    var result [][]int// 存放结果集
-    backTree(n,k,1,&track,&result)
-    return result
+    res, path = make([][]int, 0), make([]int, 0, k)
+    dfs(k, n, 1, 0)
+    return res
 }
-func backTree(n,k,startIndex int,track *[]int,result *[][]int){
-    if len(*track)==k{
-        var sum int
-        tmp:=make([]int,k)
-        for k,v:=range *track{
-            sum+=v
-            tmp[k]=v
-        }
-        if sum==n{
-            *result=append(*result,tmp)
+
+func dfs(k, n int, start int, sum int) {
+    if len(path) == k {
+        if sum == n {
+            tmp := make([]int, k)
+            copy(tmp, path)
+            res = append(res, tmp)
         }
         return
     }
-    for i:=startIndex;i<=9-(k-len(*track))+1;i++{//减枝（k-len(*track)表示还剩多少个可填充的元素）
-        *track=append(*track,i)//记录路径
-        backTree(n,k,i+1,track,result)//递归
-        *track=(*track)[:len(*track)-1]//回溯
+    for i := start; i <= 9; i++ {
+        if sum + i > n || 9-i+1 < k-len(path) {
+            break
+        }
+        path = append(path, i)
+        dfs(k, n, i+1, sum+i)
+        path = path[:len(path)-1]
     }
 }
 ```