Leetcode 155(Java)

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.

我的思路是用两个栈来实现这道题目，一个栈是正常的push，pop操作，另一个栈做成递减的，即每个元素放入时都要判断是否比当前栈头小，小的或相等的才可以放入。这样getMin时，返回min栈的栈头即可。注意相等的也应当放入，因为数字可以重复，要避免出现top了一次以后重复数字就都消失的局面。另外要注意的是，Stack不能用基本类型初始化，在用Integer包装类初始化以后，pop()中的条件比较应该要用.equals()方法而不能用==了，AC码如下：

public class MinStack {    /** initialize your data structure here. */    Stack<Integer> stack;    Stack<Integer> minStack;    public MinStack() {        stack = new Stack<Integer>();        minStack = new Stack<Integer>();    }    public void push(int x) {        stack.push(x);        if(minStack.isEmpty()||(!stack.isEmpty() && x<=minStack.peek())){            minStack.push(x);        }    }    public void pop() {        if(stack.isEmpty())return;        //注意相等值的判断        if(stack.peek().equals(minStack.peek())){            minStack.pop();        }        stack.pop();    }    public int top() {        return stack.peek();    }    public int getMin() {        return minStack.peek();    }}