LeetCode 155. Min Stack

题目：
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.

Example:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); –> Returns -3.
minStack.pop();
minStack.top(); –> Returns 0.
minStack.getMin(); –> Returns -2.

思路：
设置一个正常的栈s1,push、pop、top操作都和一般的栈一样，再设置一个用来返回最小值的栈mins，定义当栈mins为空时，或者x<=mins.top()时，x也入栈mins。在pop时，如果栈s和栈mins的top相等，则两个同时pop,否则只pop栈s。

代码：

class MinStack {public:    /** initialize your data structure here. */    MinStack() {    }    void push(int x) {        s1.push(x);        if(mins.empty()||x<=mins.top()){//定义当栈mins为空时，或者x<=mins.top()时，x也入栈mins            mins.push(x);        }    }    void pop() {        if(s1.top()==mins.top()){//如果栈s和栈mins的top相等，则两个同时pop,否则只pop栈s            mins.pop();        }        s1.pop();    }     int top() {        return s1.top();    }    int getMin() {        return mins.top();    }private:    stack<int> s1;//正常的栈s1,push、pop、top操作都和一般的栈一样    stack<int> mins;//返回最小值的栈mins};/** * Your MinStack object will be instantiated and called as such: * MinStack obj = new MinStack(); * obj.push(x); * obj.pop(); * int param_3 = obj.top(); * int param_4 = obj.getMin(); */