Merge branch 'main' of github.com:tkzzzzzz6/Algorithm_beginner_learning_notes

Author: tkzzzzzz6

acwing/problems/.cpp

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+/*
+ * @acwing app=acwing.cn id=45 lang=C++
+ *
+ * 47. 二叉树中和为某一值的路径
+ */
+
+// @acwing code start
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+class Solution {
+public:
+  vector<vector<int>> res;
+  vector<int> tracker;
+  vector<vector<int>> findPath(TreeNode *root, int sum) {
+    dfs(root, sum);
+    return res;
+  }
+
+  void dfs(TreeNode *root, int sum) {
+    if (!root)
+      return;
+    sum -= root->val;
+    tracker.push_back(root->val);
+    if (!root->right && !root->left && !sum) {
+      res.push_back(tracker);
+    }
+    dfs(root->left, sum);
+    dfs(root->right, sum);
+    tracker.pop_back();
+  }
+};
+// @acwing code end

acwing/problems/47.二叉树中和为某一值的路径.cpp

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+/*
+ * @acwing app=acwing.cn id=89 lang=C++
+ *
+ * 48. 复杂链表的复刻
+ */
+
+// @acwing code start
+
+/**
+ * Definition for singly-linked list with a random pointer.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next, *random;
+ *     ListNode(int x) : val(x), next(NULL), random(NULL) {}
+ * };
+ */
+class Solution {
+   public:
+    ListNode *copyRandomList(ListNode *head) {
+        if (!head)
+            return head;
+
+        // 1.在每个原节点后面复制新节点
+        auto p = head;
+        while (p) {
+            auto np = new ListNode(p->val);
+            np->next = p->next;
+            p->next = np;
+            p = np->next;
+        }
+
+        // 2.设置新节点的 random 指针
+        p = head;
+        while (p) {
+            if (p->random) {
+                p->next->random = p->random->next;
+            }
+            p = p->next->next;
+        }
+
+        // 3.拆分链表
+        auto dummy = new ListNode(-1);
+        auto cur = dummy;
+        p = head;
+        while (p) {
+            cur->next = p->next;
+            cur = cur->next;
+            p->next = p->next->next;
+            p = p->next;
+        }
+
+        return dummy->next;
+    }
+};
+
+// @acwing code end

acwing/problems/48.复杂链表的复刻.cpp

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+/*
+ * @Author: tkzzzzzz6
+ * @Date: 2026-04-25 11:33:20
+ * @LastEditors: tkzzzzzz6
+ * @LastEditTime: 2026-04-25 11:36:00
+ */
+/*
+ * @acwing app=acwing.cn id=87 lang=C++
+ *
+ * 49. 二叉搜索树与双向链表
+ */
+
+// @acwing code start
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+#include <utility>
+using namespace std;
+class Solution {
+   public:
+    TreeNode *convert(TreeNode *root) {
+        if (!root)
+            return nullptr;
+        auto res = dfs(root);
+        return res.first;
+    }
+
+    pair<TreeNode *, TreeNode *> dfs(TreeNode *root) {
+        if (!root->left && !root->right)
+            return {root, root};
+        if (root->left && root->right) {
+            auto lslide = dfs(root->left), rslide = dfs(root->right);
+            lslide.second->right = root, root->left = lslide.second;
+            rslide.first->left = root, root->right = rslide.first;
+            return {lslide.first, rslide.second};
+        }
+        if (root->left) {
+            auto lslide = dfs(root->left);
+            lslide.second->right = root, root->left = lslide.second;
+            return {lslide.first, root};
+        }
+        if (root->right) {
+            auto rslide = dfs(root->right);
+            rslide.first->left = root, root->right = rslide.first;
+            return {root, rslide.second};
+        }
+
+        return {nullptr, nullptr};
+    }
+};
+// @acwing code end

acwing/problems/49.二叉搜索树与双向链表.cpp

@@ -0,0 +1,62 @@
+/*
+ * @acwing app=acwing.cn id=46 lang=C++
+ *
+ * 50. 序列化二叉树
+ */
+
+// @acwing code start
+
+#include <sstream>
+#include <string>
+
+using namespace std;
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+class Solution {
+   public:
+    // Encodes a tree to a single string.
+    string serialize(TreeNode *root) {
+        string s;
+        dfs_s(root, s);
+        return s;
+    }
+
+    void dfs_s(TreeNode *root, string &s) {
+        if (!root) {
+            s += "n ";
+            return;
+        } else {
+            s += to_string(root->val) + ' ';
+        }
+        dfs_s(root->left, s);
+        dfs_s(root->right, s);
+    }
+
+    // Decodes your encoded data to tree.
+    TreeNode *deserialize(string data) {
+        stringstream ss(data);
+        return dfs_d(ss);
+    }
+
+    TreeNode *dfs_d(stringstream &ss) {
+        string s;
+        ss >> s;
+        if (s == "n")
+            return nullptr;
+
+        auto p = new TreeNode(stoi(s));
+        p->left = dfs_d(ss);
+        p->right = dfs_d(ss);
+
+        return p;
+    }
+};
+// @acwing code end

acwing/problems/50.序列化二叉树.cpp

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+/*
+ * @acwing app=acwing.cn id=47 lang=C++
+ *
+ * 51. 数字排列
+ */
+
+// @acwing code start
+
+#include <vector>
+class Solution {
+public:
+  vector<vector<int>> res;
+  vector<int> path;
+  vector<vector<int>> permutation(vector<int> &nums) {
+    path.resize(nums.size());
+    sort(nums.begin(), nums.end());
+    dfs(nums, 0, 0, 0);
+    return res;
+  }
+  void dfs(vector<int> &nums, int u, int start, int state) {
+    if (u == nums.size()) {
+      res.push_back(path);
+      return;
+    }
+    if (!u || nums[u] != nums[u - 1])
+      start = 0;
+    for (int i = start; i < nums.size(); ++i) {
+      if (!(state >> i & 1)) {
+        path[i] = nums[u];
+        dfs(nums, u + 1, i + 1, state + (1 << i));
+      }
+    }
+  }
+};
+// @acwing code end

acwing/problems/51.数字排列.cpp

@@ -0,0 +1,24 @@
+/*
+ * @acwing app=acwing.cn id=48 lang=C++
+ *
+ * 52. 数组中出现次数超过一半的数字
+ */
+
+// @acwing code start
+
+class Solution {
+public:
+  int moreThanHalfNum_Solution(vector<int> &nums) {
+    int cnt = 0, val = -1;
+    for (int i = 0; i < nums.size(); ++i) {
+      if (cnt <= 0)
+        val = nums[i];
+      if (nums[i] == val)
+        cnt++;
+      else
+        cnt--;
+    }
+    return val;
+  }
+};
+// @acwing code end

acwing/problems/52.数组中出现次数超过一半的数字.cpp

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+/*
+ * @acwing app=acwing.cn id=49 lang=C++
+ *
+ * 53. 最小的k个数
+ */
+
+// @acwing code start
+
+#include <algorithm>
+#include <queue>
+#include <vector>
+class Solution {
+public:
+  vector<int> getLeastNumbers_Solution(vector<int> input, int k) {
+    std::priority_queue<int> heap;
+    std::vector<int> res;
+
+    for (auto x : input) {
+      heap.push(x);
+      if (heap.size() > k)
+        heap.pop();
+    }
+
+    while (!heap.empty()) {
+      res.push_back(heap.top());
+      heap.pop();
+    }
+    std::reverse(res.begin(), res.end());
+    return res;
+  }
+};
+// @acwing code endnt

acwing/problems/53.最小的k个数.cpp

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+/*
+ * @acwing app=acwing.cn id=88 lang=C++
+ *
+ * 54. 数据流中的中位数
+ */
+
+// @acwing code start
+
+#include <functional>
+#include <vector>
+class Solution {
+   public:
+    priority_queue<int> max_heap;                                      // 大顶堆，存储较小的一半
+    priority_queue<int, std::vector<int>, std::greater<int>> min_heap; // 小顶堆，存储较大的一半
+    void insert(int num) {
+        max_heap.push(num);
+        if (min_heap.size() && max_heap.top() > min_heap.top()) {
+            auto maxv = max_heap.top(), minv = min_heap.top();
+            max_heap.pop(), min_heap.pop();
+            max_heap.push(minv), min_heap.push(maxv);
+        }
+        if (max_heap.size() > min_heap.size() + 1) {
+            min_heap.push(max_heap.top());
+            max_heap.pop();
+        }
+    }
+
+    double getMedian() {
+        if (max_heap.size() + min_heap.size() & 1) {
+            return max_heap.top();
+        } else {
+            return (max_heap.top() + min_heap.top()) / 2.0;
+        }
+    }
+};
+// @acwing code end

acwing/problems/54.数据流中的中位数.cpp

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+/*
+ * @acwing app=acwing.cn id=50 lang=C++
+ *
+ * 55. 连续子数组的最大和
+ */
+
+// @acwing code start
+
+#include <climits>
+class Solution {
+   public:
+    int maxSubArray(vector<int> &nums) {
+        int res = INT_MIN, s = 0;
+        for (auto x : nums) {
+            if (s < 0)
+                s = 0;
+            s += x;
+            res = max(res, s);
+        }
+        return res;
+    }
+};
+// @acwing code end