bool find132pattern(vector<int>& nums) {if(nums.size()<3)return false; stack<int> s;int second=INT_MIN;for(int i=nums.size()-1;i>=0;i--){if(nums[i]<second)return true;while(!s.empty()&&nums[i]>s.top()){second = s.top();s.pop();}s.push(nums[i]);}return false;}

int largestRectangleArea(vector<int>& h) {h.push_back(0);int res = 0;vector<int> pos;for(int i=0;i<h.size();i++){while(!pos.empty()&&h[i]<=h[pos.back()]){int cur = h[pos.back()];pos.pop_back();//出栈int sidx = pos.empty() ? -1:pos.back();res = max(res, cur*(i-sidx-1));}pos.push_back(i);}return res;}

    int maximalRectangle(vector<vector<char>>& m) {if(m.empty()) return 0;vector<int> h(m[0].size(),0);int res = 0;for(int i=0;i<m.size();i++){for(int j=0;j<m[0].size();j++){if(m[i][j]=='0') h[j] = 0;else h[j]++;}res = max(res,largestRectangleArea(h));} return res;}int largestRectangleArea(vector<int>& h) {h.push_back(0);int res = 0;vector<int> pos;for(int i=0;i<h.size();i++){while(!pos.empty()&&h[i]<=h[pos.back()]){int cur = h[pos.back()];pos.pop_back();int sidx = pos.empty() ? -1:pos.back();res = max(res, cur*(i-sidx-1));}pos.push_back(i);}return res;}

    bool isValidSerialization(string pre) {char tmp;bool isNum = false;pre += ',';vector<char> s;for(auto tmp:pre){if(tmp=='#'){while(!s.empty()&&s.back()=='#'){s.pop_back();if(s.empty()||s.back()=='#')return false;//case“###”
                    s.pop_back();}s.push_back('#');}else if(tmp==','){if(isNum)s.push_back('n');isNum = false;}else{isNum = true;}}return s.back()=='#'&&s.size()==1;   }

//思路一：
vector<int> postorderTraversal(TreeNode* r) {vector<int> v;if (!r) return v;stack<TreeNode*> s;s.push(r);while(!s.empty()){TreeNode* cur = s.top();v.insert(v.begin(),cur->val);s.pop();if(cur->left)s.push(cur->left);if(cur->right)s.push(cur->right);}return v;}//思路二：
vector<int> postorderTraversal(TreeNode* r) {vector<int> v;if (!r) return v;stack<TreeNode*> s;TreeNode* pre;s.push(r);while(!s.empty()){TreeNode* cur = s.top();if((!cur->left&&!cur->right)||(pre == cur->left&&!cur->right)||(pre&&pre==cur->right)){v.push_back(cur->val);s.pop();pre = cur;}else{if(cur->right)s.push(cur->right);if(cur->left)s.push(cur->left);}}return v;}
//思路三：vector<int> postorderTraversal(TreeNode* r) {vector<int> v;if (!r) return v;stack<TreeNode*> s;TreeNode* p = r;do{while(p){     //0,最开始将左节点全部入栈
                s.push(p);p=p->left;}TreeNode* pre = NULL;while(!s.empty()){p = s.top();if(p->right==pre){//1,访问过、或者右子树为空才出栈v.push_back(p->val);pre = p;s.pop();}else{//2,没有访问过、跳出继续将p->right的一路左子树入栈，执行0p = p->right;break;}}}while(!s.empty());return v;}

 vector<int> postorderTraversal(TreeNode* r) {vector<int> v;if (!r) return v;stack<TreeNode*> s;s.push(r);//pre表示后序序列中节点的前驱//最开始的pre不能为空，要将pre和cur的左右子节点进行比较//当访问最左边节点时，如果pre=NULL且右节点==NULL,这一点就不能够入栈TreeNode *pre =new TreeNode(0);while(!s.empty()){TreeNode* cur = s.top();//有左节点，且左右节点都还没访问过，入栈if(cur->left&&pre!=cur->left&&pre!=cur->right){s.push(cur->left);//有右节点，且没有访问过，而左节点访问过}else if(cur->right&&pre!=cur->right&&(!cur->left||cur->left == pre)){s.push(cur->right);}else{//其他情况，由于栈后进先出，栈顶出栈即可，这时候pre就指向curv.push_back(cur->val);pre = cur;s.pop();}}return v;}

#include<iostream>
#include<cstring>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<functional>
using namespace std;unordered_map<char, function<int  (int, int )>> op_map = {{ '+' , [] (int  a, int  b) { return a + b; } },{ '-' , [] (int  a, int  b) { return a - b; } },{ '*' , [] (int  a, int  b) { return a * b; } },{ '/' , [] (int  a, int  b) { return a / b; } }};
unordered_map<char, int> priority = {{'+',1},{'-',1},{'*',2},{'/',2},{'(',0}
};/* 利用两个栈进行模拟计算 */
int Compute(string s)
{stack<char> opstk;    //操作符栈stack<int> numstk;  //操作数栈int pos = 0;s = "(" + s + ")";int cur;while(pos < s.length()){if (s[pos] == '('){opstk.push('(');if(s[++pos]=='-')cur = 0;//负号必定出现在'('的后面，变成减法}else if (s[pos] == ')'){//括号内计算得到的cur不需要入栈while (opstk.top() != '('){cur = op_map[opstk.top()](numstk.top(), cur);//只需拿cur与栈顶数字进行计算
                numstk.pop();opstk.pop();}opstk.pop();  //删除'('pos++;}else if (s[pos] >= '0' && s[pos] <= '9'){int integer = 0;while (s[pos] >= '0' && s[pos] <= '9'){integer = integer*10 + (s[pos++] - '0');}cur = integer;}else{while(priority[s[pos]] <= priority[opstk.top()]){cur = op_map[opstk.top()](numstk.top(), cur);numstk.pop();opstk.pop();}numstk.push(cur);opstk.push(s[pos++]);}}return cur;
}int main()
{string s = "-2+3*(-2+3*4)-(-(-2))";cout << "结果为：" << Compute(s) << endl;//输出结果为26return 0;
}

#include <queue>
#include<iostream>
#include<stack>
#include<cstdio>
using namespace std;/*定义状态*/
typedef struct Statute{char left,right,mid;int n;Statute(){};Statute(char L,char m,char r,int n):left(L),mid(m),right(r),n(n){};
}Statute;/*用栈*/
void HanNouWei(Statute s){stack<Statute> stk;stk.push(s);int cnt = 0;while(!stk.empty()){Statute cur = stk.top();stk.pop();if(cur.n==1){cout << " from "<< cur.left << " to " << cur.right  << endl;}else{stk.push(Statute(cur.mid,cur.left,cur.right,cur.n-1));//此处并不能够直接输出结果，而应该将这一步的状态放入栈中//cout << " from "<< cur.left << " to " << cur.right  << endl;stk.push(Statute(cur.left,cur.mid,cur.right,1));stk.push(Statute(cur.left,cur.right,cur.mid,cur.n-1));}}}/*递归*/
void hanoi(Statute s){Statute cur = s;if(cur.n == 1) {cout << " from "<< cur.left << " to " << cur.right  << endl;return;} else {hanoi(Statute(cur.left,cur.right,cur.mid,cur.n-1));cout << " from "<< cur.left << " to " << cur.right  << endl;hanoi(Statute(cur.mid,cur.left,cur.right,cur.n-1));}}int main(){Statute s = Statute('A','B','C',4);HanNouWei(s);cout << "==============================="<< endl;hanoi(s);return 0;
}