1047. 删除字符串中的所有相邻重复项 - 力扣（LeetCode）

class Solution {
public:string removeDuplicates(string s) {string stack;for(char& ch:s){if(stack.size()>0&&ch==stack.back()){stack.pop_back();}else{stack.push_back(ch);}}return stack;}
};

844. 比较含退格的字符串 - 力扣（LeetCode）

class Solution {
public:bool backspaceCompare(string s, string t) {string tmp_s;string tmp_t;for(char& ch:s){if(ch=='#'){if(tmp_s.size()>0)tmp_s.pop_back();}else{tmp_s.push_back(ch);}}for(char& ch:t){if(ch=='#'){if(tmp_t.size()>0)tmp_t.pop_back();}else{tmp_t.push_back(ch);}}if(tmp_s==tmp_t) return true;return false;}
};

227. 基本计算器 II - 力扣（LeetCode）

class Solution {
public:int calculate(string s) {vector<int> stack;//栈内不存在加减乘除符号，符号遍历时用op保存，如果有括号的话，需要另一个栈，然后加减符号可以用正负代替插入。int i=0; int n=s.size();char op='+';while(i<n){if(s[i]==' ') i++;else if(0+'0'<=s[i]&&s[i]<=0+'9'){//提取数字int tmp=s[i++]-'0';while(0+'0'<=s[i]&&s[i]<=0+'9') tmp+=tmp*10+s[i++]-'0';cout<<"op: "<<op<<endl;cout<<tmp<<endl;if(op=='+') stack.push_back(tmp);else if(op=='-') stack.push_back(-tmp);else if(op=='*'){int tmp1=stack.back();//必须在push前先popstack.pop_back();stack.push_back(tmp1*tmp);// stack.pop_back();}else if(op=='/'){int tmp1=stack.back();stack.pop_back();stack.push_back(tmp1/tmp);} for(int num:stack){cout<<num<<" ";}cout<<endl;}else {// cout<<"op: "<<op<<endl;op=s[i++];}}int ret=0;for(int& num:stack){// cout<<num<<endl;ret+=num;}return ret;}
};

946. 验证栈序列 - 力扣（LeetCode）

class Solution {
public:bool validateStackSequences(vector<int>& pushed, vector<int>& popped){stack<int> st;int i=0; int n=popped.size();for(int &val : pushed){st.push(val);while(st.size()&&st.top()==popped[i]){st.pop();i++;}}return st.size()==0;}
};