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Min Stack

Source

Implement a stack with min() function,
which will return the smallest number in the stack.

It should support push, pop and min operation all in O(1) cost.

Example
Operations: push(1), pop(), push(2), push(3), min(), push(1), min() Return: 1, 2, 1

Note
min operation will never be called if there is no number in the stack

题解

『最小』栈，要求在栈的基础上实现可以在 O(1)O(1) 的时间内找出最小值，一般这种 O(1)O(1)的实现往往就是哈希表或者哈希表的变体，这里简单起见可以另外克隆一个栈用以跟踪当前栈的最小值。

Java

public class Solution {
    public Solution() {
        stack1 = new Stack<Integer>();
        stack2 = new Stack<Integer>();
    }

    public void push(int number) {
        stack1.push(number);
        if (stack2.empty()) {
            stack2.push(number);
        } else {
            stack2.push(Math.min(number, stack2.peek()));
        }
    }

    public int pop() {
        stack2.pop();
        return stack1.pop();
    }

    public int min() {
        return stack2.peek();
    }

    private Stack<Integer> stack1; // original stack
    private Stack<Integer> stack2; // min stack
}

源码分析

取最小栈的栈顶值时需要先判断是否为空栈(而不仅是 null)。

复杂度分析

均为 O(1)O(1).