leetcode 155. Min Stack 最小栈的实现

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.

就是使用两个栈来实现本题描述的最小栈的实现，一个栈正常使用，另一个栈保存最小值，直接上代码吧。

代码如下：

import java.util.Arrays;
import java.util.Stack;

public class MinStack 
{
    private Stack stk;
    private Stack min;

    public MinStack() 
    {
        stk = new Stack<>();
        min = new Stack<>();
    }

    public void push(int x) 
    {
        stk.push(x);
        if(min.isEmpty() || x <= min.peek())
            min.push(x);
    }

    /*
     * 注意这里必须要使用equals，因为Integer是一个整数对象
     * */
    public void pop() 
    {
        if(stk.peek().equals(min.peek()))
        {
            stk.pop();
            min.pop();
        }else
            stk.pop();
    }

    public int top() 
    {
        return stk.peek();
    }

    public int getMin() 
    {
        return min.peek();

    }
}

/**
 * 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();
 */