s0047

include/s0047_permutations_ii.hpp

@@ -0,0 +1,14 @@
+#ifndef S0047_PERMUTATIONS_II_HPP
+#define S0047_PERMUTATIONS_II_HPP
+
+#include <algorithm>
+#include <vector>
+
+using namespace std;
+
+class S0047 {
+ public:
+  vector<vector<int>> permuteUnique(vector<int>& nums);
+};
+
+#endif

notes/src/permute.md

@@ -1,3 +1,83 @@
 # 排列问题
 
 ## [46. 全排列](https://leetcode.cn/problems/permutations/)
+
+## [47. 全排列 II](https://leetcode.cn/problems/permutations-ii/)
+
+和 s0046 相比就加了去重。有两种去重思路，一个是用哈希表记录每一层的使用情况。另一种是排序 + 判断，后者性能更好所以优先选择后者。
+
+哈希表法：
+
+```cpp
+#include "s0047_permutations_ii.hpp"
+
+void permuteUniqueDFS(vector<int> &path, vector<vector<int>> &result,
+                      vector<bool> &used, vector<int> &nums) {
+  int len = nums.size();
+  // 终止条件
+  if (path.size() == len) {
+    result.push_back(path);
+    return;
+  }
+
+  // 创建一个哈希表用来记录当前层中使用过的元素
+  unordered_map<int, bool> map;
+
+  // 开始迭代
+  for (int i{0}; i < len; ++i) {
+    // 如果当前元素在树枝或树层使用过，则跳过
+    if (used[i] || map.count(nums[i]) == 1) continue;
+    // 否则处理当前节点
+    map[nums[i]] = true;
+    path.push_back(nums[i]);
+    used[i] = true;
+    permuteUniqueDFS(path, result, used, nums);
+    used[i] = false;
+    path.pop_back();
+  }
+}
+
+vector<vector<int>> S0047::permuteUnique(vector<int> &nums) {
+  vector<int> path{};
+  vector<vector<int>> result{};
+  vector<bool> used(nums.size(), false);
+  permuteUniqueDFS(path, result, used, nums);
+  return result;
+}
+```
+
+排序 + 判断:
+
+```cpp
+void permuteUniqueDFS(vector<int> &path, vector<vector<int>> &result,
+                      vector<bool> &used, vector<int> &nums, int startIndex) {
+  int len = nums.size();
+  // 终止条件
+  if (path.size() == len) {
+    result.push_back(path);
+    return;
+  }
+
+  // 开始迭代
+  for (int i{0}; i < len; ++i) {
+    // 如果当前元素在树层使用过，则跳过
+    if (used[i] || (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false))
+      continue;
+    // 否则处理当前节点
+    path.push_back(nums[i]);
+    used[i] = true;
+    permuteUniqueDFS(path, result, used, nums, i + 1);
+    used[i] = false;
+    path.pop_back();
+  }
+}
+
+vector<vector<int>> S0047::permuteUnique(vector<int> &nums) {
+  vector<int> path{};
+  vector<vector<int>> result{};
+  vector<bool> used(nums.size(), false);
+  sort(nums.begin(), nums.end());
+  permuteUniqueDFS(path, result, used, nums, 0);
+  return result;
+}
+```

src/s0047_permutations_ii.cpp

@@ -0,0 +1,33 @@
+#include "s0047_permutations_ii.hpp"
+
+void permuteUniqueDFS(vector<int> &path, vector<vector<int>> &result,
+                      vector<bool> &used, vector<int> &nums, int startIndex) {
+  int len = nums.size();
+  // 终止条件
+  if (path.size() == len) {
+    result.push_back(path);
+    return;
+  }
+
+  // 开始迭代
+  for (int i{0}; i < len; ++i) {
+    // 如果当前元素在树层使用过，则跳过
+    if (used[i] || (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false))
+      continue;
+    // 否则处理当前节点
+    path.push_back(nums[i]);
+    used[i] = true;
+    permuteUniqueDFS(path, result, used, nums, i + 1);
+    used[i] = false;
+    path.pop_back();
+  }
+}
+
+vector<vector<int>> S0047::permuteUnique(vector<int> &nums) {
+  vector<int> path{};
+  vector<vector<int>> result{};
+  vector<bool> used(nums.size(), false);
+  sort(nums.begin(), nums.end());
+  permuteUniqueDFS(path, result, used, nums, 0);
+  return result;
+}