【Leetcode-Easy-155】 Min Stack-优快云博客

本文链接：https://blog.youkuaiyun.com/liyazhou0215/article/details/78285412

本文介绍了一种能够在常数时间内支持push、pop、top及获取最小元素操作的MinStack设计。通过使用两个栈来实现这一目标，即一个数据栈用于存放所有元素，另一个最小值栈用于跟踪当前最小值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

【Leetcode-Easy-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.

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.

思路

双栈

程序

class MinStack {

    private LinkedList<Integer> dataStack = null;
    private LinkedList<Integer> minStack = null; 

    /** initialize your data structure here. */
    public MinStack() {
        dataStack = new LinkedList<>();
        minStack = new LinkedList<>();
    }

    public void push(int x) {
        dataStack.push(x);
        if (minStack.isEmpty()) minStack.push(x);
        else                    minStack.push(Math.min(minStack.peek(), x));
    }

    public void pop() {
        if(!minStack.isEmpty()) minStack.pop();
        if(!dataStack.isEmpty()) dataStack.pop();
    }

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

    public int getMin() {
        return minStack.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();
 */

声明：题目来自Leetcode.