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.

链接： http://leetcode.com/problems/min-stack/

2/18/2017, Java

注意！我们其实可以直接用stack不用ArrayList的！

Integer 和int是不相等的，第14行如果是Integer x = ...那么15行永远不相等！

public class MinStack {
    ArrayList<Integer> list = new ArrayList<Integer>();
    ArrayList<Integer> min = new ArrayList<Integer>();

    /** initialize your data structure here. */
    public MinStack() {}
    
    public void push(int x) {
        list.add(x);
        if (min.size() == 0 || x <= min.get(min.size() - 1)) min.add(x);
    }

    public void pop() {
        int x = list.remove(list.size() - 1);
        if (x == min.get(min.size() - 1)) min.remove(min.size() - 1);
    }
    
    public int top() {
        return list.get(list.size() - 1);
    }
    
    public int getMin() {
        return min.get(min.size() - 1);
    }
}

4/16/2017

BB电面准备

Java Stack里没有top()，但是有peek()

public class MinStack {
    Stack<Integer> s;
    Stack<Integer> ms;
    /** initialize your data structure here. */
    public MinStack() {
        s = new Stack<Integer>();
        ms = new Stack<Integer>();
    }
    
    public void push(int x) {
        s.push(x);
        if (ms.empty() || ms.peek() >= x) ms.push(x);
    }
    
    public void pop() {
        int t = s.peek();
        if (t <= ms.peek()) ms.pop();
        s.pop();
    }
    
    public int top() {
        return s.peek();
    }
    
    public int getMin() {
        return ms.peek();
    }
}