From 0d6a3a751373f0c5a1ffeac11efbd125d2cbc5d9 Mon Sep 17 00:00:00 2001
From: Sainnhe Park <i@sainnhe.dev>
Date: Thu, 2 Feb 2023 11:33:45 +0800
Subject: [PATCH] s0040

---
 include/s0040_combination_sum_ii.hpp | 14 ++++++++++
 notes/src/combinations.md            | 27 ++++++++++++++++++-
 src/s0040_combination_sum_ii.cpp     | 39 ++++++++++++++++++++++++++++
 3 files changed, 79 insertions(+), 1 deletion(-)
 create mode 100644 include/s0040_combination_sum_ii.hpp
 create mode 100644 src/s0040_combination_sum_ii.cpp

diff --git a/include/s0040_combination_sum_ii.hpp b/include/s0040_combination_sum_ii.hpp
new file mode 100644
index 0000000..35212aa
--- /dev/null
+++ b/include/s0040_combination_sum_ii.hpp
@@ -0,0 +1,14 @@
+#ifndef S0040_COMBINATION_SUM_II_HPP
+#define S0040_COMBINATION_SUM_II_HPP
+
+#include <algorithm>
+#include <vector>
+
+using namespace std;
+
+class S0040 {
+ public:
+  vector<vector<int>> combinationSum2(vector<int>& candidates, int target);
+};
+
+#endif
diff --git a/notes/src/combinations.md b/notes/src/combinations.md
index c51c7a3..7d8a0e9 100644
--- a/notes/src/combinations.md
+++ b/notes/src/combinations.md
@@ -69,6 +69,31 @@ void combineDFS(int n, int k, int begin, vector<int> &path,
 }
 ```
 
+## [216. 组合总和 III](https://leetcode.cn/problems/combination-sum-iii/)
+
 ## [39. 组合总和](https://leetcode.cn/problems/combination-sum/)
 
-## [216. 组合总和 III](https://leetcode.cn/problems/combination-sum-iii/)
+## [40. 组合总和 II](https://leetcode.cn/problems/combination-sum-ii/)
+
+最难的一个组合总和，因为 `candidates` 有重复元素，而要求最终结果不能重复。
+
+e.g. 1
+
+```text
+Input: candidates = [10,1,2,7,6,1,5], target = 8
+Output:
+[
+[1,1,6],
+[1,2,5],
+[1,7],
+[2,6]
+]
+```
+
+如果你只是单纯地在 s0039 的基础上在下一次递归中将 `startIndex` 设为 `i + 1` 那么最终结果就会出现两个 `[1, 2, 5]`。
+
+如果你直接排除 `candidates[i] == candidates[i - 1]` 的情形，那么最终结果就没有 `[1, 1, 6]`。
+
+正确的逻辑应该是如果 `candidates[i] == candidates[i - 1]` 且 `candidates[i - 1]` 使用过，则剪枝。
+
+怎么判断 `candidates[i - 1]` 是否使用过呢？我们创建一个 `vector<bool> used` 用来记录每个元素是否使用过。
diff --git a/src/s0040_combination_sum_ii.cpp b/src/s0040_combination_sum_ii.cpp
new file mode 100644
index 0000000..604cb5b
--- /dev/null
+++ b/src/s0040_combination_sum_ii.cpp
@@ -0,0 +1,39 @@
+#include "s0040_combination_sum_ii.hpp"
+
+void combinationSum2DFS(vector<int> &candidates, int target, int startIndex,
+                        vector<int> &path, int sum, vector<bool> &used,
+                        vector<vector<int>> &result) {
+  // 结束条件：总和等于 target 。不存在总和大于 target 的情况，因为已经被剪枝了
+  if (sum == target) {
+    result.push_back(path);
+    return;
+  }
+
+  // 开始迭代
+  int size = candidates.size();
+  for (int i = startIndex; i < size; ++i) {
+    // 剪枝，当现在节点的 sum 已经超过了 target，就没必要继续迭代了
+    if (sum + candidates[i] > target) break;
+    // 剪枝，要对同一树层使用过的元素进行跳过
+    if (i > 0 && candidates[i] == candidates[i - 1] && used[i - 1] == false)
+      continue;
+    path.push_back(candidates[i]);
+    used[i] = true;
+    combinationSum2DFS(candidates, target, i + 1, path, sum + candidates[i],
+                       used, result);
+    used[i] = false;
+    path.pop_back();
+  }
+}
+
+vector<vector<int>> S0040::combinationSum2(vector<int> &candidates,
+                                           int target) {
+  // 对 candidates 进行升序排序，这是为了进行剪枝
+  sort(candidates.begin(), candidates.end());
+  // 初始化
+  vector<int> path{};
+  vector<vector<int>> result{};
+  vector<bool> used(candidates.size(), false);
+  combinationSum2DFS(candidates, target, 0, path, 0, used, result);
+  return result;
+}