feat: 添加多个算法题的解决方案及相关代码实现

Author: tkzzzzzz6

acwing/problems/.cpp

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+/*
+* @acwing app=acwing.cn id=38 lang=C++
+*
+* 
+*/
+
+// @acwing code start
+
+
+// @acwing code end

acwing/problems/36.合并两个排序的链表.cpp

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+/*
+ * @Author: tkzzzzzz6
+ * @Date: 2026-04-22 08:01:54
+ * @LastEditors: tkzzzzzz6
+ * @LastEditTime: 2026-04-22 08:09:06
+ */
+/*
+* @acwing app=acwing.cn id=34 lang=C++
+*
+* 36. 合并两个排序的链表
+*/
+
+// @acwing code start
+
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode* merge(ListNode* l1, ListNode* l2) {
+        auto dummy = new ListNode(0);
+        auto cur = dummy;
+
+        while(l1 && l2){
+            if(l1->val > l2->val){
+                cur->next = l2;
+                l2 = l2->next;
+            }else{
+                cur->next = l1;
+                l1 = l1->next;
+            }
+            cur = cur->next;
+        }
+
+        if(l1){
+            cur->next = l1;
+        }else{
+            cur->next = l2;
+        }
+
+        return dummy->next;
+    }
+};
+// @acwing code end

acwing/problems/37.树的子结构.cpp

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+/*
+* @acwing app=acwing.cn id=35 lang=C++
+*
+* 37. 树的子结构
+*/
+
+// @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:
+    bool hasSubtree(TreeNode* pRoot1, TreeNode* pRoot2) {
+        if(!pRoot1 || !pRoot2)return false;
+        if(isSame(pRoot1,pRoot2))return true;
+        return hasSubtree(pRoot1->left,pRoot2) || hasSubtree(pRoot1->right,pRoot2);
+    }
+
+    bool isSame(TreeNode* root1,TreeNode* root2){
+        if(!root2)return true;
+        if(!root1 || root1->val != root2->val)return false;
+        return isSame(root1->left,root2->left) && isSame(root1->right,root2->right);
+    }
+};
+// @acwing code end

acwing/problems/38.二叉树的镜像.cpp

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+/*
+* @acwing app=acwing.cn id=37 lang=C++
+*
+* 38. 二叉树的镜像
+*/
+
+// @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:
+    void mirror(TreeNode* root) {
+        if(!root)return;
+        swap(root->left,root->right);
+        mirror(root->left);
+        mirror(root->right);
+    }
+};
+// @acwing code end

acwing/problems/39.对称的二叉树.cpp

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+/*
+ * @Author: tkzzzzzz6
+ * @Date: 2026-04-22 10:58:49
+ * @LastEditors: tkzzzzzz6
+ * @LastEditTime: 2026-04-22 11:01:05
+ */
+/*
+* @acwing app=acwing.cn id=38 lang=C++
+*
+* 39. 对称的二叉树
+*/
+
+// @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:
+    bool isSymmetric(TreeNode* root) {
+        return !root || dfs(root->left,root->right);
+    }
+
+    bool dfs(TreeNode* p,TreeNode* q){
+        if(!p || !q)return !p && !q;
+
+        return (p->val == q->val) && dfs(p->left,q->right) && dfs(p->right,q->left);
+    }
+};
+// @acwing code end

acwing/problems/40.顺时针打印矩阵.cpp

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+/*
+* @acwing app=acwing.cn id=39 lang=C++
+*
+* 40. 顺时针打印矩阵
+*/
+
+// @acwing code start
+
+class Solution {
+public:
+    vector<int> printMatrix(vector<vector<int> > matrix) {
+        if(matrix.empty() || matrix[0].empty())return vector<int>();
+        
+        vector<int> res;
+        int n = matrix.size(),m = matrix[0].size();
+        int dx[4] = {0,1,0,-1},dy[4] = {1,0,-1,0};
+
+        int d = 0,x = 0,y = 0;
+        vector<vector<bool>> vis(n,vector<bool>(m));
+
+        for(int k = 0;k<n*m;++k){
+            res.push_back(matrix[x][y]);
+            vis[x][y] = true;
+
+            int a = x + dx[d];
+            int b = y + dy[d];
+
+            if(a<0 || a >= n || b < 0 || b >= m || vis[a][b]){
+                d = (d+1) %4;
+                a = x+dx[d],b = y + dy[d];
+            }
+            x = a,y = b;
+        }
+        return res;
+    }
+};
+// @acwing code end

acwing/problems/41.包含min函数的栈.cpp

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+/*
+* @acwing app=acwing.cn id=90 lang=C++
+*
+* 41. 包含min函数的栈
+*/
+
+// @acwing code start
+
+class MinStack {
+public:
+    /** initialize your data structure here. */
+    stack<int> st,stmin;
+    MinStack() {
+        
+    }
+    
+    void push(int x) {
+        st.push(x);
+        if(stmin.empty()){
+            stmin.push(x);
+        }else{
+            stmin.push(min(x,stmin.top()));
+        }
+    }
+    
+    void pop() {
+        if(st.empty())return;
+        st.pop();
+        stmin.pop();
+    }
+    
+    int top() {
+        return st.empty() ? -1 : st.top();
+    }
+    
+    int getMin() {
+        return stmin.empty() ? -1 : stmin.top();
+    }
+};
+
+/**
+ * Your MinStack object will be instantiated and called as such:
+ * MinStack obj = new MinStack();
+ * obj.push(x);
+ * obj.pop();
+ * int param_3 = obj.top();
+ * int param_4 = obj.getMin();
+ */
+// @acwing code end

acwing/problems/42.栈的压入、弹出序列.cpp

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+/*
+* @acwing app=acwing.cn id=40 lang=C++
+*
+* 42. 栈的压入、弹出序列
+*/
+
+// @acwing code start
+
+class Solution {
+public:
+    stack<int> st;
+    bool isPopOrder(vector<int> pushV,vector<int> popV) {
+        if(pushV.size() != popV.size())return false;
+
+        int j = 0;
+        for(int x:pushV){
+            st.push(x);
+            while(!st.empty() && j < popV.size() && st.top() == popV[j]){
+                st.pop();
+                ++j;
+            }
+        }
+
+        return j == popV.size();
+    }
+};
+// @acwing code end

acwing/problems/43.不分行从上往下打印二叉树.cpp

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+/*
+* @acwing app=acwing.cn id=41 lang=C++
+*
+* 43. 不分行从上往下打印二叉树
+*/
+
+// @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<int> printFromTopToBottom(TreeNode* root) {
+        vector<int> res;
+        if(!root)return res;
+
+        queue<TreeNode*> q;
+        q.push(root);
+
+        while(q.size()){
+            auto t = q.front();
+            q.pop();
+            res.push_back(t->val);
+            if(t->left)q.push(t->left);
+            if(t->right)q.push(t->right);
+        }
+
+        return res;
+    }
+};
+// @acwing code end

acwing/problems/44.分行从上往下打印二叉树.cpp

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+/*
+* @acwing app=acwing.cn id=42 lang=C++
+*
+* 44. 分行从上往下打印二叉树
+*/
+
+// @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>> printFromTopToBottom(TreeNode* root) {
+        vector<vector<int>> res;
+        if(!root)return res;
+
+        queue<TreeNode*> q;
+        q.push(root);
+        q.push(nullptr);
+
+        vector<int> level;
+        while(q.size()){
+            auto t = q.front();
+            q.pop();
+
+            if(t == nullptr){
+                res.push_back(level);
+                level.clear();
+                if(!q.empty()){
+                    q.push(nullptr);
+                }
+            }else{
+                level.push_back(t->val);
+                if(t->left)q.push(t->left);
+                if(t->right)q.push(t->right);
+            }
+        }
+
+        return res;
+    }
+};
+// @acwing code end