[C++ brushing notes] stack classic OJ (minimum stack, stack push pop sequence, reverse Polish expression evaluation)

 Record the classic OJ and wrong questions in the learning process

(1) 155. Minimal Stack - LeetCode

class MinStack {
public:
    MinStack() {

    }
    
    void push(int val) {
        // 只要是压栈，先将元素保存到_st中
        _st.push(val);
        
        // 如果x小于_minst中栈顶的元素，将x再压入_minst中
        if(_minst.empty() || val <= _minst.top())
            _minst.push(val);
    }
    
    void pop() {
        // 如果_minst栈顶的元素等于出栈的元素，_min顶的元素要移除
        if(_st.top() == _minst.top())
            _minst.pop();

        _st.pop();
    }
    
    int top() {
        return _st.top();
    }
    
    int getMin() {
        return _minst.top();
    }
    
    // 保存栈中的元素
    stack<int> _st;

    // 保存栈的最小值
    stack<int> _minst;
};

(2) Push-in and pop-up sequences of the stack

class Solution {
public:
    bool IsPopOrder(vector<int> pushV,vector<int> popV) {
    stack<int> st;
    size_t popi = 0;

    for(auto e : pushV)
    {
        //入栈
        st.push(e);

        //跟出栈序列相比，相同就出
        while(!st.empty() && st.top() == popV[popi])
        {
            st.pop();
            popi++;
        }
    }

    //栈为非空则报错
    return st.empty();

    }
};

(3) 150. Reverse Polish Expression Evaluation - LeetCode

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        stack<int> st;

        for(auto str : tokens)
        {
            //遇到运算符就运算，没有遇到就压栈
            if(str=="+" || str=="-" || str=="*" || str=="/")
            {
                int right = st.top();
                st.pop();
                int left = st.top();
                st.pop();

                switch(str[0])//这里不可以使用str
                {
                    case '+':
                        st.push(left+right);
                        break;
                    case '-':
                        st.push(left-right);
                        break;
                    case '*':
                        st.push(left*right);
                        break;
                    case '/':
                        st.push(left/right);
                        break;
                }
            }
            else
            {
                //stoi可以将字符转化成整形
                st.push(stoi(str));
            }
        }

        return st.top();
    }
};

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