diff --git a/problems/0047.全排列II.md b/problems/0047.全排列II.md
index c809c62d..d9fe8f35 100644
--- a/problems/0047.全排列II.md
+++ b/problems/0047.全排列II.md
@@ -49,7 +49,7 @@
 
 **一般来说：组合问题和排列问题是在树形结构的叶子节点上收集结果，而子集问题就是取树上所有节点的结果**。
 
-在[46.全排列](https://programmercarl.com/0046.全排列.html)中已经详解讲解了排列问题的写法，在[40.组合总和II](https://programmercarl.com/0040.组合总和II.html) 、[90.子集II](https://programmercarl.com/0090.子集II.html)中详细讲解的去重的写法，所以这次我就不用回溯三部曲分析了，直接给出代码，如下：
+在[46.全排列](https://programmercarl.com/0046.全排列.html)中已经详细讲解了排列问题的写法，在[40.组合总和II](https://programmercarl.com/0040.组合总和II.html) 、[90.子集II](https://programmercarl.com/0090.子集II.html)中详细讲解了去重的写法，所以这次我就不用回溯三部曲分析了，直接给出代码，如下：
 
 ## C++代码
 
@@ -225,33 +225,37 @@ class Solution:
 ### Go 
 
 ```go
-var res [][]int
-func permute(nums []int) [][]int {
-	res = [][]int{}
-	backTrack(nums,len(nums),[]int{})
-	return res
+var (
+    res [][]int
+    path  []int
+    st    []bool   // state的缩写
+)
+func permuteUnique(nums []int) [][]int {
+    res, path = make([][]int, 0), make([]int, 0, len(nums))
+    st = make([]bool, len(nums))
+    sort.Ints(nums)
+    dfs(nums, 0)
+    return res
 }
-func backTrack(nums []int,numsLen int,path []int)  {
-	if len(nums)==0{
-		p:=make([]int,len(path))
-		copy(p,path)
-		res = append(res,p)
-	}
-	used := [21]int{}//跟前一题唯一的区别，同一层不使用重复的数。关于used的思想carl在递增子序列那一题中提到过
-	for i:=0;i<numsLen;i++{
-		if used[nums[i]+10]==1{
-			continue
-		}
-		cur:=nums[i]
-		path = append(path,cur)
-		used[nums[i]+10]=1
-		nums = append(nums[:i],nums[i+1:]...)
-		backTrack(nums,len(nums),path)
-		nums = append(nums[:i],append([]int{cur},nums[i:]...)...)
-		path = path[:len(path)-1]
-
-	}
 
+func dfs(nums []int, cur int) {
+    if cur == len(nums) {
+        tmp := make([]int, len(path))
+        copy(tmp, path)
+        res = append(res, tmp)
+    }
+    for i := 0; i < len(nums); i++ {
+        if i != 0 && nums[i] == nums[i-1] && !st[i-1] {  // 去重，用st来判别是深度还是广度
+            continue
+        }
+        if !st[i] {
+            path = append(path, nums[i])
+            st[i] = true
+            dfs(nums, cur + 1)
+            st[i] = false
+            path = path[:len(path)-1]
+        }
+    }
 }
 ```