随笔分类 - leetcode
摘要：``` class Solution { public: vector> res; vector path; bool st[10]; void dfs(vector nums,int u) { if(u==nums.size()) { res.push_back(path); return; }
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摘要：``` class Solution { public: unordered_map cnt; vector> res; vector path; vector> subsetsWithDup(vector& nums) { for(auto i:nums) cnt[i]++; dfs(-10);/
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摘要：``` class Solution { public: vector> res; vector> combinationSum2(vector& candidates, int target) { sort(candidates.begin(),candidates.end()); dfs(can
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摘要：``` class Solution { public: vector> res; vector> combinationSum(vector& candidates, int target) { dfs(candidates,0,target); return res; } vector path
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摘要：```c class Solution { public: void dfs(TreeNode* root,int &sum)//从大到小遍历所有节点 { if(!root) return; dfs(root->right,sum); sum+=root->val; root->val=sum; d
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摘要：``` class Solution { public: vector res; int cnt=0; int find(TreeNode* root,int val)//返回当前子树值为val的个数 { if(!root) return 0; return find(root->left,val)
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摘要：```c class Solution { public: vector path1,path2; bool dfs(TreeNode* root,TreeNode* p,vector& path) { if(!root) return false; if(root==p||dfs(root->le
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摘要：```c class Solution { public: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { if(p->valval&&q->valval) return lowestCommonA
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摘要：```c class Solution { public: vector res; unordered_map hashmap;//记录每一个子树出现的次数 string dfs(TreeNode* root) { if(!root) return ""; string str=""; str+=t
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摘要：# 思路1 ``` class Solution { public: void flatten(TreeNode* root) { while(root) { auto p=root->left; if(p)//找到左儿子的右链 { while(p->right) p=p->right; //将右链
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摘要：``` class Solution { public: vector dfs(TreeNode* root)//依次返回是否是二叉搜索树，最大值最小值 { vector res{1,root->val,root->val}; if(root->left) { auto l=dfs(root->le
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摘要：``` class Solution { public: int countNodes(TreeNode* root) { if(!root) return 0; auto l=root->left,r=root->right; int x=1,y=1;//记录左右两边层数 while(l) l=l
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摘要：``` class Solution { public: vector dfs(int l,int r)//返回以n为根节点的所有可能子树 { vector res; if(l>r) return {NULL}; for(int k=l;k left=dfs(l,k-1); vector right
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摘要：``` class Solution { public: vector> res; void bfs(TreeNode* root) { queue q; q.push(root); int cnt=0; while(!q.empty()) { vector level; int len=q.siz
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摘要：``` class Solution { public: vector> res; void bfs(TreeNode* root) { queue q; q.push(root); int last=0; while(!q.empty()) { vector level; int len=q.si
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摘要：``` /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr),
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摘要：``` class Solution { public: vector inorderTraversal(TreeNode* root) { vector res; stack st; while(root||st.size()) { while(root) { st.push(root); roo
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摘要：class Solution { public: string longestPalindrome(string s) { string res; int n=s.size(); for(int i=0;i<n;i++) { //长度是奇数 int l=i-1,r=i+1; while(l>=0&&
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摘要：可以直接分类讨论，分别枚举第一个房屋偷或不偷的情况，最后再取极值 不偷第一家 f[i] 代表前 i 个房屋，偷第 i 个, 且不偷第一家的最大值 g[i] 代表前 i 个房屋，不偷第 i 个, 且不偷第一家的最大值 偷第一家 f[i] 代表前 i 个房屋，偷第 i 个, 且偷第一家的最大值 g[i]
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摘要：class Solution { public: int maxProfit(vector<int>& prices) { int n=prices.size(); vector<int> f(n+1),g(n+1); //预处理f，f[i]表示前i天交易的最大值 for(int i=1,min_p
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