class Solution {
public:
string replaceSpace(string s){
int len = s.size();
string g;for(int i =0; i < len; i++){
if(s[i]==' '){
g +="%20";continue;}
g += s[i];}return g;}};
class Solution {
public:intstrToInt(string str){
int len = str.size();int i =0;int flag =1;//标记正负longlong num =0;//返回转化后的数字while(str[i]==' ')//去掉前导空格{
i++;}if(str[i]=='+'||str[i]=='-'){
if(str[i]=='-')
flag =-flag;
i++;}while(i < len && str[i]>='0'&& str[i]<='9'){
num = num*10+(str[i]-'0')*flag;
i++;if(flag ==-1&& num < INT_MIN ){
return INT_MIN;}if(flag ==1&& num > INT_MAX ){
return INT_MAX;}}if(i != len)//没有遍历完就是错误，直接返回当前的值{
return num;}return num;}};
class Solution {
public:voidexchange(vector<int>&nums,int len){
int* ret =(int*)malloc(sizeof(int)*(len +1));int j =0;for(int i =0; i < len; i++){
if((nums[i]&1)==1){
ret[j++]= nums[i];}}for(int i =0; i < len; i++){
if((nums[i]&1)==0){
ret[j++]= nums[i];}}for(int i =0; i < len; i++){
nums[i]= ret[i];}}
vector<int>exchange(vector<int>& nums){
int len = nums.size();exchange(nums,len);return nums;}};
class Solution {
public:
vector<int>twoSum(vector<int>& nums,int target){
int len = nums.size();
unordered_map<int,int>hash;
vector<int>ret;for(int i =0; i < len; i++){
int num = target - nums[i];if(hash.count(num)){
ret.push_back(num);
ret.push_back(nums[i]);break;}
hash.insert({
nums[i],i});}return ret;}};
class Solution {
public:
vector<int>maxSlidingWindow(vector<int>& nums,int k){
int len = nums.size();
deque<int>q;
vector<int>ret;for(int i =0; i < nums.size(); i++){
if(!q.empty()&& i - k +1> q[0]) q.pop_front();//下标小于滑动窗口则头删//形成单调递增的队列while(!q.empty()&& nums[q[q.size()-1]]<= nums[i]) q.pop_back();
q.push_back(i);if(i >= k -1){
ret.push_back(nums[q[0]]);}}return ret;}};
class CQueue {
public:CQueue(){
}voidappendTail(int value){
p_push.push(value);}intdeleteHead(){
if(p_push.empty()&& p_pop.empty()){
return-1;}if(!p_pop.empty()){
int top = p_pop.top();
p_pop.pop();return top;}else{
while(!p_push.empty()){
p_pop.push(p_push.top());
p_push.pop();}}int top = p_pop.top();
p_pop.pop();return top;}
private:
stack<int>p_push;
stack<int>p_pop;};/**
* Your CQueue object will be instantiated and called as such:
* CQueue* obj = new CQueue();
* obj->appendTail(value);
* int param_2 = obj->deleteHead();
*/
class MinStack {
stack<int>x_push;
stack<int>min_push;
public:/** initialize your data structure here. */MinStack(){
min_push.push(INT_MAX);}voidpush(int x){
x_push.push(x);
min_push.push(std::min(min_push.top(),x));//每次插入插最小的值进去，和最小值的栈顶比较。}voidpop(){
x_push.pop();
min_push.pop();}inttop(){
return x_push.top();}intmin(){
return min_push.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->min();
*/
class MaxQueue {
public:
deque<int>val;
deque<int>vmax;MaxQueue(){
}intmax_value(){
if(vmax.empty()){
return-1;}return vmax.front();}voidpush_back(int value){
while((!vmax.empty())&&(vmax.back()< value))//单调递减的队列{
vmax.pop_back();}
val.push_back(value);
vmax.push_back(value);}intpop_front(){
if(val.empty()){
return-1;}int ans = vmax.front();if(ans == val.front())//只有和最大值相等才需要删除，跟滑动窗口差不多{
vmax.pop_front();}int front = val.front();
val.pop_front();return front;}};/**
* Your MaxQueue object will be instantiated and called as such:
* MaxQueue* obj = new MaxQueue();
* int param_1 = obj->max_value();
* obj->push_back(value);
* int param_3 = obj->pop_front();
*/
class Solution {
public:intfindRepeatNumber(vector<int>& nums){
unordered_map<int,int>hash;int k =0;for(int i =0; i < nums.size(); i++){
if(hash.count(nums[i]))//hash判断该数有没有出现过{
k = i;return nums[k];}
hash.insert({
nums[i],i});}return nums[k];}};
class Solution {
public:intmissingNumber(vector<int>& nums){
int n = nums.size();int sum =0;for(int i =1; i <= n; i++){
sum += i;}for(int i =0; i < n; i++){
sum -= nums[i];}return sum;}};
class Solution {
public:intminArray(vector<int>& numbers){
//这是一个左旋数组int n = numbers.size();//二分查找法int left =0,right = n -1;while(left < right){
int mid = left + right >>1;if(numbers[mid]> numbers[right]) left = mid +1;elseif(numbers[mid]< numbers[right]) right = mid;//当出现nums[m] = nums[j] 时，一定有区间[i, m] 内所有元素相等 或 区间[m, j] 内所有元素相等（或两者皆满 足）。对于寻找此类数组的最小值问题，可直接放弃二分查找，而使用线性查找替代。else right--;}return numbers[left];}};
class Solution {
public:
vector<vector<int>>levelOrder(TreeNode* root){
if(root == nullptr)return{
};
queue<TreeNode*>q;
vector<vector<int>>ret;
q.push(root);while(!q.empty()){
int num = q.size();
ret.push_back(vector<int>());for(int i =0; i < num; i++){
auto node = q.front();
q.pop();
ret.back().push_back(node->val);//在尾部进行尾插，因为插入了二维空vectorif(node->left) q.push(node->left);if(node->right) q.push(node->right);}}int k = ret.size();for(int i =0; i < k; i++){
if(i %2!=0)//奇数反转{
reverse(ret[i].begin(),ret[i].end());}}return ret;}};
class Solution {
public://DFS算法
bool IsEmpty(vector<vector<char>>& board, vector<vector<int>>&visited,int rows,int cols,string word,int k){
int a = board.size(),b = board[0].size();if(board[rows][cols]!= word[k])return false;//如果字符不同为假elseif(k == word.size()-1)return true;//匹配成功返回
visited[rows][cols]= true;
bool result = false;for(int i =0; i <4; i++){
int x = rows + dx[i],y = cols + dy[i];if(x >=0&& x < a && y >=0&& y < b)//判断边界{
if(!visited[x][y])//判断是否访问过{
bool flag =IsEmpty(board, visited, x, y, word, k +1);if(flag)return true;//如果成功直接返回}}}
visited[rows][cols]= false;//跳出循环要取消标记return result;}
bool exist(vector<vector<char>>& board, string word){
int row = board.size(),col = board[0].size();
vector<vector<int>>visited(row,vector<int>(col));
bool flag = false;for(int i =0; i < row; i++)for(int j =0; j < col; j++){
flag =IsEmpty(board,visited,i,j,word,0);if(flag)return true;}return false;}
private:int dx[4]={
0,0,-1,1};int dy[4]={
1,-1,0,0};};
//我写的dfs
class Solution {
public:intget(int x){
int sum =0;while(x >0){
sum += x %10;
x /=10;}return sum;}voidcheck(vector<vector<int>>&ans,int row,int col,int k,int&res,int sum,int o){
if(sum > k)return;
res++;int a = ans.size(),b = ans[0].size();
ans[row][col]= true;//走过就不要走了for(int i =0; i <2; i++){
int x = row + dx[i], y = col + dy[i];if(x >=0&& x < a && y >=0&& y < b &&!ans[x][y]){
int q =get(x), p =get(y);int sum1 = p + q;check(ans,x,y,k,res,sum1,o+1);}}}intmovingCount(int m,int n,int k){
if(k ==0)return1;
vector<vector<int>>ans(m,vector<int>(n));int res =0;//表示走格子的最大值int o =1;int sum =0;//记录数位和check(ans,0,0,k,res,sum,o);return res;}
private:int dx[4]={
1,0};//优化成下和右int dy[4]={
0,1};};//官方题解bfs
class Solution {
// 计算 x 的数位之和intget(int x){
int res=0;for(; x; x /=10){
res += x %10;}return res;}
public:intmovingCount(int m,int n,int k){
if(!k)return1;
queue<pair<int,int>> Q;// 向右和向下的方向数组int dx[2]={
0,1};int dy[2]={
1,0};
vector<vector<int>>vis(m, vector<int>(n,0));
Q.push(make_pair(0,0));
vis[0][0]=1;int ans =1;while(!Q.empty()){
auto[x, y]= Q.front();
Q.pop();for(int i =0; i <2;++i){
int tx = dx[i]+ x;int ty = dy[i]+ y;if(tx <0|| tx >= m || ty <0|| ty >= n
|| vis[tx][ty]||get(tx)+get(ty)> k)continue;
Q.push(make_pair(tx, ty));
vis[tx][ty]=1;
ans++;}}return ans;}};
class Codec {
public:voidrserialize(TreeNode * root,string & ret){
//前序遍历转为字符串if(root == nullptr){
ret +="#,";return;}
ret +=to_string(root->val);//字符转换函数
ret +=",";rserialize(root->left,ret);rserialize(root->right,ret);}// Encodes a tree to a single string.
string serialize(TreeNode* root){
string ret;rserialize(root,ret);return ret;}// Decodes your encoded data to tree.
TreeNode*rdeserialize(list<string>&dateArray){
if(dateArray.front()=="#"){
dateArray.erase(dateArray.begin());return nullptr;}
TreeNode * root = new TreeNode(stoi(dateArray.front()));//将字符串转为数字
dateArray.erase(dateArray.begin());//插入后删除
root->left =rdeserialize(dateArray);
root->right =rdeserialize(dateArray);return root;}
TreeNode*deserialize(string data){
list<string>dateArray;//用链表存字符串
string str;for(auto& c : data){
if(c ==','){
dateArray.push_back(str);
str.clear();}else{
str.push_back(c);}}if(!str.empty())//如果后面有节点就需要插入在清空{
dateArray.push_back(str);
str.clear();}returnrdeserialize(dateArray);}};// Your Codec object will be instantiated and called as such:// Codec codec;// codec.deserialize(codec.serialize(root));
class Solution {
public:
vector<int>printNumbers(int n){
vector<int>ret;if(n ==0)return ret;int k =1;while(n--){
k *=10;}for(int i =1; i < k; i++){
ret.push_back(i);}return ret;}};
class Solution {
public:doublePow(double x,longlong n){
if(n ==0)return1.0;double y =Pow(x,n /2);//快速幂算法//如果n为偶数就是返回的结果的平方，否则就是返回的结果多乘xreturn n %2==0? y * y : y * y * x;}doublemyPow(double x,int n){
if(n ==0)return1.0;longlong N = n;//防止越界return N >=0?Pow(x,N):1/Pow(x,-N);}};
class Solution {
public:intevalRPN(vector<string>& tokens){
stack<int>st;for(auto& ch : tokens){
if(ch =="+"|| ch =="-"|| ch =="*"|| ch =="/"){
int right = st.top();
st.pop();int left = st.top();
st.pop();char str = ch[0];int num;switch(str){
case'+':
num = left + right;
st.push(num);break;case'-':
num = left - right;
st.push(num);break;case'*':
num = left * right;
st.push(num);break;case'/':
num = left / right;
st.push(num);break;}}else{
st.push(stoi(ch));}}return st.top();}};
class Solution {
public:intmaxProfit(vector<int>& prices){
int len = prices.size();int mini =0x3f3f3f3f;//买股票的最小值int maxi =0;//卖股票的最大值for(auto price : prices){
maxi =max(maxi,price - mini);
mini =min(mini,price);}return maxi;}};
class Solution {
public:intnthUglyNumber(int n){
priority_queue<double,vector<double>,greater<double>>q;double ans =1;for(int i =1; i < n; i++){
q.push(ans*2);
q.push(ans*3);
q.push(ans*5);
ans = q.top();
q.pop();while(!q.empty()&& q.top()== ans) q.pop();}return ans;}};
class Solution {
public:
vector<int>GetNum(vector<int>&nums,int sz){
vector<int>ret;int k =0;for(int i =0; i <31; i++){
if((sz >> i)&1==1){
k = i;break;}}int a =0, b =0;for(int i =0; i < nums.size(); i++){
if(((nums[i]>> k)&1)==0){
a ^= nums[i];}else{
b ^= nums[i];}}
ret.push_back(a);
ret.push_back(b);return ret;}
vector<int>singleNumbers(vector<int>& nums){
vector<int>ret;int sz =0;for(int i =0; i < nums.size(); i++){
sz ^= nums[i];}returnGetNum(nums,sz);}};