leetcode MinStack

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

1, 记得stack的几个函数,isEmpty(),不能用null, equals,还有读stack最上面的数,但是不取出来的peek()

2,解法,用另外一个stack来计算最小的,最上面的永远是最小的

class MinStack {
    Stack<Integer> sta=new Stack<Integer>();
    Stack<Integer> minsta=new Stack<Integer>();
    public void push(int x) {
       sta.push(x);
       if(minsta.isEmpty()||x<=minsta.peek()){
           minsta.push(x);
       }
    }
    public void pop() {       
        if(sta.peek().equals(minsta.peek())){
            minsta.pop();
        }
        sta.pop();
    }

    public int top() {
        return sta.peek();
    }

    public int getMin() {
        return minsta.peek();
    }
}

 

posted @ 2014-11-11 05:04  lilyfindjob  阅读(215)  评论(0编辑  收藏  举报