LeetCode——Min Stack

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.

原题链接：https://oj.leetcode.com/problems/min-stack/

取出栈中的最小元素需要常数时间。

可以用两个栈来实现，其中一个栈存储的是全部的元素，而另一个栈存储的是前一个栈中的最小元素，这样每次更新后一个栈即可。

public class MinStack {
	private List<Integer> stack = new ArrayList<Integer>();
	private List<Integer> minstack = new ArrayList<Integer>();
	public void push(int x) {
		stack.add(x);
		if(minstack.isEmpty() || minstack.get(minstack.size()-1) >= x)
			minstack.add(x);
	}

	public void pop() {
		if(stack.isEmpty())
			return;
		int tmp = stack.remove(stack.size()-1);
		if(!minstack.isEmpty() && tmp == minstack.get(minstack.size()-1))
			minstack.remove(minstack.size()-1);
	}

	public int top() {
		if(!stack.isEmpty())
			return stack.get(stack.size()-1);
		return -1;
	}

	public int getMin() {
		if(!minstack.isEmpty())
			return minstack.get(minstack.size()-1);
		return -1;
	}
}