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
      
        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; } }