LeetCode——Min Stack

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. 原題鏈接：https://oj.leetcode.com/problems/min-stack/ 取出棧中的最小元素需要常數時間。 可以用兩個棧來實現，其中一個棧存儲的是全部的元素，而另一個棧存儲的是前一個棧中的最小元素，這樣每次更新後一個棧即可。

  public class MinStack {
  	private List stack = new ArrayList();
  	private List minstack = new ArrayList();
  	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;
  	}
  }