【leetcode】155

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

• push(x) -- Push element x onto stack.
• pop() -- Removes the element on top of the stack.
• top() -- Get the top element.
• getMin() -- Retrieve the minimum element in the stack.

Solution:  若只维护一个栈stk，在每次getMin的时候检索stk需要两倍的临时空间，倒两次，故用两个栈，一个stk,一个min

class MinStack {
public:
    void push(int x) {
        stk.push(x);
        if(min.empty()||x<=min.top())min.push(x);       //if的两个判断条件顺序不能替换，否则stk添加第一个元素后getMin出错
    }
    void pop() {
        if(stk.top()==min.top()){
            stk.pop();
            min.pop();
        }else
            stk.pop();
    }
    int top() {
        return stk.top();
    }
    int getMin() {
        return min.top();
    }
private:
    stack<int> stk;
    stack<int> min;
};