Leetcode155——Min Stack

文章作者：Tyan
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1. 问题描述

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.

2. 求解

方法一
主要是模拟写一个最小栈。要注意push时可能会输入null。需要使用双栈实现，一个保存数据，一个保存最小值。由于随着数据出栈，最小值是不断变化的，因此需要一个最小值栈来保存最小值。方法一是最小值栈与普通栈大小不等的情况。

class MinStack {
    private Stack<Integer> stack = new Stack<>();
    private Stack<Integer> minStack = new Stack<>();

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

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

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

    public int getMin() {
        return minStack.peek();        
    }
}

方法二
最小值栈与存储元素的栈大小始终相等的情况

class MinStack {
    private Stack<Integer> stack = new Stack<Integer>();
    private Stack<Integer> minStack = new Stack<Integer>();

    public void push(int x) {
        if(minStack.isEmpty() || x <= minStack.peek()) {
            minStack.push(x);
        }else {
            minStack.push(minStack.peek());
        }
        stack.push(x);
    }

    public void pop() {
        minStack.pop();
        stack.pop();
    }

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

    public int getMin() {
        return minStack.peek();        
    }
}