L_Aster的专栏

小区便利店正在促销，用 numExchange 个空酒瓶可以兑换一瓶新酒。你购入了 numBottles 瓶酒。如果喝掉了酒瓶中的酒，那么酒瓶就会变成空的。请你计算 最多 能喝到多少瓶酒。示例 1：输入：numBottles = 9, numExchange = 3输出：13解释：你可以用 3 个空酒瓶兑换 1 瓶酒。所以最多能喝到 9 + 3 + 1 = 13 瓶酒。示例2输入：numBottles = 15, numExchange = 4输出：19解释：你可以用 4 个空酒瓶兑

给你一棵树（即，一个连通的无环无向图），这棵树由编号从 0 到 n - 1 的 n 个节点组成，且恰好有 n - 1 条 edges 。树的根节点为节点 0 ，树上的每一个节点都有一个标签，也就是字符串 labels 中的一个小写字符（编号为 i 的 节点的标签就是 labels[i] ）边数组 edges 以 edges[i] = [ai, bi] 的形式给出，该格式表示节点 ai 和 bi 之间存在一条边。返回一个大小为 n 的数组，其中 ans[i] 表示第 i 个节点的子树中与节点 i 标签相

func(arr, l, r){ if(r < l){ return -1000000000 } ans = arr[l] for(i = l + 1; i <= r; i++){ ans = ans & arr[i] } return ans}Winston 构造了一个如上所示的函数 func 。他有一个整数数组 arr 和一个整数 target ，他想找到让 |func(arr, l, r) - target| 最小的 l 和 r 。请你返回 |func(

200. 岛屿的个数class Solution {public: bool isvalid(int x,int y,int m,int n){ return x&gt;=0&amp;&amp;x&lt;m&amp;&amp;y&gt;=0&amp;&amp;y&lt;n; } int numIslands(vector&lt;vector&lt;ch...

111. 二叉树的最小深度class Solution {public: int minDepth(TreeNode* root) { if(!root) return 0; int mindepth=0x7fffffff; mDhelp(root,1,mindepth); return mindepth; }...

101. 对称二叉树class Solution {public: bool isSymmetric(TreeNode* root) { if(root==NULL) return true; return sychelp(root-&gt;left,root-&gt;right); } bool sychelp(TreeNode ...

107. 二叉树的层次遍历 IIclass Solution {public: vector&lt;vector&lt;int&gt;&gt; levelOrderBottom(TreeNode* root) { vector&lt;vector&lt;int&gt;&gt; res; lOBhelp(res,root,0); reve...

684. 冗余连接 并查集class Solution {public: vector&lt;int&gt; findRedundantConnection(vector&lt;vector&lt;int&gt;&gt;&amp; edges) { int n=edges.size(); vector&lt;int&gt; parents(n+1);...

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.Note: You can only move either down or right at a...

199. 二叉树的右视图class Solution {public: vector&lt;int&gt; rightSideView(TreeNode* root) { vector&lt;int&gt; res; rSVhelp(root,res,0); return res; } void rSVhelp(Tree...

515. 在每个树行中找最大值class Solution {public: vector&lt;int&gt; largestValues(TreeNode* root) { vector&lt;int&gt; res; lVhelp(root,res,0); return res; } void lVhelp(Tre...

207. 课程表class Solution {public: bool canFinish(int numCourses, vector&lt;pair&lt;int, int&gt;&gt;&amp; prerequisites) { vector&lt;int&gt; indegree(numCourses,0); vector&lt;vecto...

210. 课程表 IIclass Solution {public: vector&lt;int&gt; findOrder(int numCourses, vector&lt;pair&lt;int, int&gt;&gt;&amp; prerequisites) { vector&lt;int&gt; indegree(numCourses,0),res; ...

169. 求众数 虽然用map记录也能过，但是明显其复杂度高，一个o(n)的解法（貌似是某年408算法题）class Solution {public: int majorityElement(vector&lt;int&gt;&amp; nums) { int res=0,counts=0; for(int x:nums){ ...

513. 找树左下角的值class Solution {public: int findBottomLeftValue(TreeNode* root) { int maxdep=0,res=0; help(root,1,maxdep,res); return res; } void help(TreeNode *r,in...

491. 递增子序列class Solution {public: vector&lt;vector&lt;int&gt;&gt; findSubsequences(vector&lt;int&gt;&amp; nums) { set&lt;vector&lt;int&gt;&gt; res; vector&lt;int&gt; rp; ...

310. Minimum Height Trees 从叶节点开始，依次删除他们，最后剩下1或者2个节点的就是树根，有点类似拓扑排序的感觉 参考：https://leetcode.com/problems/minimum-height-trees/discuss/76055/Share-some-thoughtsclass Solution {public: vector&lt;i...

1. 两数之和class Solution {public: vector&lt;int&gt; twoSum(vector&lt;int&gt;&amp; nums, int target) { unordered_map&lt;int,int&gt; mp; for(size_t i=0;i&lt;nums.size();++i){ ...

Given n non-negative integers representing the histogram’s bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.Above is a histogram where width of each...

On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of ...

void pHelp(vector> &res,vector &rp,vector &nums,unordered_map &vis){ if(rp.size()==nums.size()) { res.push_back(rp); return; } for(unsigned int i=0;i<nums.size();++i) { if(vis[num

int findCircleNum(vector<vector<int>>& M){ int n=M.size(); vector<int> parents(n,0); for(int i=0;i<n;++i) { parents[i]=i; } function<int(int,vector<int>&)>finds=[&finds

原题链接：Median of Two Sorted Arrays 题目大意是，给你两个已经排好序的数组，找到这两个数组的中位数 比如：Example 1:nums1 = [1, 3]nums2 = [2]The median is 2.0又比如：Example 2:nums1 = [1, 2]nums2 = [3, 4]The median is (2 + 3)/2 = 2.5问题可以延

int subarraySum2(vector& nums, int k)//use hash{ int sum=0,cnt=0; unordered_map hash; hash[0]=1; for(int x:nums) { sum+=x; cnt+=hash[sum-k]; ++hash[sum]; } return cnt;}

回溯void cbhelp(vector> &res,vector &v,int target,int i,vector recp){ if(target<0) return; else if(target==0) { res.push_back(recp); return; } for(unsigned int k=i;k<v.size();++k) { recp.

注意和39题区别，当前层值相同则跳过回溯法void cb2help(vector> &res,vector &v,int target,unsigned int i,vector recp){ if(target<0) return; else if(target==0) { res.push_back(recp); return; } for(unsigned int

题目链接：点击打开链接dpint minSteps(int n){ vector dp(n+1,1024); dp[1]=0; for(int i=2;i<=n;++i) { for(int k=1;k*k<=i;++k) { if(i%k==0) dp[i]=min(min(dp[k]+i/k,dp[i/k]+k),dp[i]); } } retu

题解来自点击打开链接，vector findDuplicates(vector& nums){ vector res; for(unsigned int i=0;i<nums.size();++i) { nums[abs(nums[i])-1]=-nums[abs(nums[i])-1]; if(nums[abs(nums[i])-1]>0) res.push_back(abs

vector twoSum(vector& nums, int target){ unordered_map hash; int n=nums.size(); for(int i=0;i<n;++i) { if(hash[target-nums[i]]) return {i,hash[target-nums[i]]-1}; hash[nums[i]]=i+1; } retur

void reorderList(ListNode* head){ if(head==NULL||head->next==NULL) return; ListNode *l1=head,*l2=head->next; while(l2&&l2->next) { l1=l1->next; l2=l2->next->next; } ListNode *h=l1-

void aveLevHelp(vector &res,vector &cnt,unsigned int lev,TreeNode *r){ if(r==NULL) return; if(lev+1>res.size()) { res.push_back(0); cnt.push_back(0); } res[lev]+=r->val; cnt[lev]++

int dbHelp(TreeNode *r,int &res){ if(!r)return 0; int lh=dbHelp(r->left,res)+1; int rh=dbHelp(r->right,res)+1; res=max(res,lh+rh-2); return max(lh,rh);}int diameterOfBinaryTree(TreeNod

void lobHelp(vector> &res,unsigned int lev,TreeNode *r){ if(r==NULL) return; if(lev+1>res.size()) { res.push_back(vector()); } res[lev].push_back(r->val); lobHelp(res,lev+1,r->left);

void wbtHelp(vector &lf,int idx,unsigned int lev,int &maxwid,TreeNode *r){ if(r==NULL) return; if(lev+1>lf.size()) lf.push_back(idx); else maxwid=max(maxwid,idx+1-lf[lev]); wbtHelp(lf,2*idx,

void zzloHelp(vector> &res,unsigned int lev,TreeNode *r){ if(!r)return; if(lev+1>res.size()) { res.push_back(vector()); } res[lev].push_back(r->val); zzloHelp(res,lev+1,r->left); zz

void ssHelp(vector> &res,vector &nums,vector &rp,unsigned int cur){ res.push_back(rp); for(unsigned int i=cur;i<nums.size();++i) { rp.push_back(nums[i]); ssHelp(res,nums,rp,i+1); rp.p

void caHelp(int &res,vector &v,int N,int cnt){ if(cnt==N+1) { ++res; return; } for(int i=1;i<=N;++i) { if(v[i]==0&&(i%cnt==0||cnt%i==0)) { v[i]=1; caHelp(res,v,N,cnt+1)