一：删除字符串中的所有相邻重复项

class Solution {
public:
    string removeDuplicates(string s) {
        stack<char> st;
        for(int i = 0; i < s.size(); i++)
        {
            char target = s[i];
            if(!st.empty() && target == st.top())
                st.pop();
            else
                st.push(s[i]);
        }
        string ret;
        while(!st.empty())
        {
            ret.insert(ret.begin(), st.top());
            st.pop();
        }
        return ret;
    }
};

二：比较含退格的字符串

class Solution {
public:
    bool backspaceCompare(string s, string t) {
        stack<char> st_s;
        stack<char> st_t;
        for(auto& es : s)
        {
            if(!st_s.empty() && es == '#')
                st_s.pop();
            else if(st_s.empty() && es == '#')
                ;
            else
                st_s.push(es);
        }
        for(auto& et : t)
        {
            if(!st_t.empty() && et == '#')
                st_t.pop();
            else if(st_t.empty() && et == '#')
                ;
            else
                st_t.push(et);
        }
        return st_s == st_t;
    }
};

三：基本计算器II

class Solution {
public:
    int calculate(string s) {
        vector<int> st;     // 使用数组来模拟栈结构
        int i = 0, n = s.size();
        char op = '+';
        while(i < n)
        {
            if(s[i] == ' ')
                i++;
            else if(s[i]>='0' && s[i] <= '9')
            {
                // 先把这个数字提取出来
                int tmp = 0;
                while(i < n && s[i] >= '0' && s[i] <= '9')
                    tmp = tmp*10 + (s[i++] - '0');
                if(op == '+')
                    st.push_back(tmp);
                else if(op == '-')
                    st.push_back(-tmp);
                else if(op == '*')
                    st.back() *= tmp;
                else
                    st.back() /= tmp;
            }
            else{
                op = s[i];
                i++;
            }
        }
        int ret = 0;
        for(auto& x : st)
            ret += x;
        return ret;

    }
};

四：字符串解码

class Solution {
public:
    string decodeString(string s) {
        stack<int> nums;
        stack<string> st;
        st.push("");
        int i = 0, n = s.size();

        while(i < n)
        {
            if(s[i] >= '0' && s[i] <= '9')
            {
                // 先把这个数字提取出来
                int tmp = 0;
                while(s[i] >= '0' && s[i] <= '9')
                {
                    tmp = tmp*10 + (s[i] - '0');
                    i++;
                }
                nums.push(tmp);
            }
            else if(s[i] == '[')
            {
                // 提取 [] 内的string
                string tmp;
                i++;
                while(s[i] >= 'a' && s[i] <= 'z')
                {
                    tmp += s[i];
                    i++;
                }
                st.push(tmp);
            }
            else if(s[i] == ']')
            {
                string tmp = st.top();
                st.pop();

                int k = nums.top();
                nums.pop();
                while(k--)
                {
                    st.top() += tmp;
                }
                i++;    // 跳过这个右括号
            }
            else
            {
                string tmp;
                while(i < n && s[i] >= 'a' && s[i] <= 'z')
                {
                    tmp += s[i];
                    i++;
                }
                st.top() += tmp;
            }
        }
        return st.top();
    }
};

五：验证栈序列

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