leetcode_题解_min stack

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.

设置2个栈，一个存放数据，一个栈用栈顶存放当前的最小值，解法如下（没有考虑栈空之类的情况，OJ上反正AC了)

<pre name="code" class="cpp">class MinStack {public:    void push(int x) {       realStack.push(x);       if(minStack.empty() || minStack.top()>=x)//注意realStack和minStack的元素个数不一定是相同的，如果用相同的那种方法，可能会MLE            minStack.push(x);    }        void pop() {         int topVal=realStack.top();            realStack.pop();            if(topVal<=minStack.top())                minStack.pop();    }    int top() {            return realStack.top();    }    int getMin() {            return minStack.top();    }private:    stack<int> realStack,minStack;};