From 940815d95fe091654919885f4ca75636ba127b5b Mon Sep 17 00:00:00 2001
From: youngyangyang04 <826123027@qq.com>
Date: Thu, 6 May 2021 09:37:40 +0800
Subject: [PATCH] =?UTF-8?q?=E8=A1=A5=E5=85=85=E5=85=B6=E4=BB=96=E8=AF=AD?=
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---
 problems/0047.全排列II.md | 73 ++++++++++++++++++++++++++++++++++++
 1 file changed, 73 insertions(+)

diff --git a/problems/0047.全排列II.md b/problems/0047.全排列II.md
index b9f157f9..952aed1f 100644
--- a/problems/0047.全排列II.md
+++ b/problems/0047.全排列II.md
@@ -147,6 +147,79 @@ if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == true) {
 
 是不是豁然开朗了！！
 
+## 其他语言版本
+
+java:
+
+```java
+class Solution {
+    //存放结果
+    List<List<Integer>> result = new ArrayList<>();
+    //暂存结果
+    List<Integer> path = new ArrayList<>();
+
+    public List<List<Integer>> permuteUnique(int[] nums) {
+        boolean[] used = new boolean[nums.length];
+        Arrays.fill(used, false);
+        Arrays.sort(nums);
+        backTrack(nums, used);
+        return result;
+    }
+
+    private void backTrack(int[] nums, boolean[] used) {
+        if (path.size() == nums.length) {
+            result.add(new ArrayList<>(path));
+            return;
+        }
+        for (int i = 0; i < nums.length; i++) {
+            // used[i - 1] == true，说明同⼀树⽀nums[i - 1]使⽤过
+            // used[i - 1] == false，说明同⼀树层nums[i - 1]使⽤过
+            // 如果同⼀树层nums[i - 1]使⽤过则直接跳过
+            if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false) {
+                continue;
+            }
+            //如果同⼀树⽀nums[i]没使⽤过开始处理
+            if (used[i] == false) {
+                used[i] = true;//标记同⼀树⽀nums[i]使⽤过，防止同一树支重复使用
+                path.add(nums[i]);
+                backTrack(nums, used);
+                path.remove(path.size() - 1);//回溯，说明同⼀树层nums[i]使⽤过，防止下一树层重复
+                used[i] = false;//回溯
+            }
+        }
+    }
+}
+```
+
+python：
+
+```python
+class Solution:
+    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
+        # res用来存放结果
+        if not nums: return []
+        res = []
+        used = [0] * len(nums)
+        def backtracking(nums, used, path):
+            # 终止条件
+            if len(path) == len(nums):
+                res.append(path.copy())
+                return
+            for i in range(len(nums)):
+                if not used[i]:
+                    if i>0 and nums[i] == nums[i-1] and not used[i-1]:
+                        continue
+                    used[i] = 1
+                    path.append(nums[i])
+                    backtracking(nums, used, path)
+                    path.pop()
+                    used[i] = 0
+        # 记得给nums排序
+        backtracking(sorted(nums),used,[])
+        return res
+```
+
+
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