07 2023 档案
摘要：``` class Solution { public: int hammingDistance(int x, int y) { int cnt=0; int t=x^y; while(t) { cnt+=(t&1); t>>=1; } return cnt; } }; ```
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摘要：``` class Solution { public: vector findDisappearedNumbers(vector& nums) { vector res; int n=nums.size(),offset=n+1; for(auto &num:nums) { int x=(num)
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摘要：``` class Solution { public: vector findAnagrams(string s, string p) { unordered_map window,hash; vector res; for(auto i:p) hash[i]++; for(int i=0,j=0
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摘要：``` class Solution { public: int subarraySum(vector& nums, int k) { int n=nums.size(),res=0; vector s(n+1,0); unordered_map hash;//记录端点i之前所有前缀和的出现情况 f
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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: struct node { int val; int pre; node* next; node(int a,int b,node* c) { val=a; pre=b; next=c; } }; void insert(node* &hea
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摘要：``` class Solution { public: vector calcEquation(vector>& equations, vector& values, vector>& queries) { unordered_set node;//记录所有节点 unordered_map> g;
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摘要：``` class Solution { public: string dfs(string s,int &idx) { string str; while(idx'0'&&s[idx]='0'&&s[idx]<='9') num+=s[idx++]; if(s[idx]=='[') { int c
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摘要：# 快排思想 - 注意，这里是倒序排序，因此应该`while(nums[i].cnt>x);` ``` class Solution { public: struct element { int val,cnt; element(int a,int b) { val=a; cnt=b; } }; v
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摘要：``` class Solution { public: vector countBits(int n) { vector f(n+1); for(int i=1;i>1]+(i&1); return f; } }; ```
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摘要：## 思路 本题有两点需要解决，如果确定括号的最少删除数量以及删除哪些括号 #### 求出最少删除数量 思路见 LeetCode 22. 括号生成 #### 已知删除数量，删除哪些括号？ 直接爆搜，枚举每个括号是否删除 - 如果当前是左括号，枚举是否删除 (删除 0 还是 1 个) - 如果当前是左
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摘要：``` class Solution { public: int findDuplicate(vector& nums) { if(nums.size()<2) return nums[0]; int n=nums.size(); int fast=0,slow=0; do { slow=nums[
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摘要：``` class Solution { public: void moveZeroes(vector& nums) { if(nums.empty()) return; int n=nums.size(); int idx=n-1; while(idx>=0&&nums[idx]==0) idx-
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摘要：``` class Solution { public: bool searchMatrix(vector>& matrix, int target) { if(matrix.empty()||matrix[0].empty()) return false; int n=matrix.size(),
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摘要：``` class Solution { public: vector maxSlidingWindow(vector& nums, int k) { deque q; vector res; for(int i=0;i=k) q.pop_front(); //插入新元素 while(q.size(
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摘要：``` class Solution { public: vector productExceptSelf(vector& nums) { vector q; int t=1; for(auto i:nums) { q.push_back(t); t*=i; } t=1; for(int i=num
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摘要：``` class Solution { public: ListNode* reverse(ListNode* head)//翻转以head为头节点的链表 { if(!head||!head->next) return head; auto a=head,b=head->next; while(b
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摘要：# 小根堆 ``` class Solution { public: int findKthLargest(vector& nums, int k) { priority_queue,greater> q; for(auto x:nums) { if(q.size()& nums, int k) {
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摘要：``` class Solution { public: bool canFinish(int n, vector>& pre) { if(pre.empty()||pre[0].empty()) return true; vector> g(n,vector(n,false)); for(auto
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摘要：``` /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x
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摘要：``` class Solution { public: bool st[310][310]; int dx[4]={0,0,-1,1},dy[4]={-1,1,0,0}; int m,n; int numIslands(vector>& g) { int res=0; n=g.size(),m=g
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摘要：``` class Solution { public: int majorityElement(vector& nums) { int cnt=1; int res=nums[0]; for(int i=1;i<nums.size();i++) { if(nums[i]==res) cnt++;
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摘要：``` /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class
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摘要：#使用双栈 ``` class MinStack { public: stack st, min_st; MinStack() { min_st.push(INT_MAX); } void push(int val) { st.push(val); int min_val=min_st.top();
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摘要：``` class Solution { public: static const int N=20010; int f[N],g[N]; int maxProduct(vector& nums) { int n=nums.size(); int res=nums[0]; f[0]=g[0]=num
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摘要：``` class Solution { public: ListNode* sortList(ListNode* head) { if(!head||!head->next) return head; ListNode* fast=head,*slow=head; while(fast->next
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摘要：``` class LRUCache { public: struct node { int key,val; node *l,*r; node(int a,int b) { l=r=NULL; key=a; val=b; } }*L,*R; unordered_map mp;//保存key和节点的
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摘要：``` /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class
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