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.

分析：

本题考察栈的操作，要求实现push,pop,top,getMin四个函数，对于push,pop,top可以直接调用java中Stack的相关方法；getMin()函数可以通过新设一个存储最小元素的栈来实现，意思就是创建两个栈，一个栈正常执行相关操作，另一个栈只存储当前最小栈元素。

代码：

import java.util.Stack;

public class minStack {
Stack<Integer> stack=new Stack<Integer>();
Stack<Integer> minStack=new Stack<Integer>();
public void push(int x) {
if(minStack.isEmpty()|| x<minStack.peek()){
minStack.push(x);
}
stack.push(x);
}

public void pop() {
if(minStack.peek().equals(stack.peek()))
minStack.pop();
stack.pop();
}

public int top() {
int top=stack.peek();
return top;
}

public int getMin() {
int min=minStack.peek();
return min;
}
public static void main(String[] args){
minStack ms=new minStack();
ms.push(-3);
int x=ms.getMin();
System.out.println("min:"+x);
}
}