155. 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.

 

Example:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> Returns -3.
minStack.pop();
minStack.top();      --> Returns 0.
minStack.getMin();   --> Returns -2.


这道题没c#的,所以用Java写的。用两个stack,一个存push的值,另外一个用来存最小值。Pop的时候,如果正常stack pop的值小于等于另一个存最小值的stack,最小值的stack也pop
public class MinStack {
    private Stack _stack;
    private Stack _stackMin;
    /** initialize your data structure here. */
    public MinStack() {
        _stack = new Stack();
        _stackMin = new Stack();
    }
    
    public void push(int x) {
        _stack.push(x);
        if(_stackMin.isEmpty() ||  (int)_stackMin.peek() >= x)
        {
            _stackMin.push(x);
        }
    }
    
    public void pop() {
        
        if(!_stack.isEmpty() &&  (int)_stack.pop() <= (int)_stackMin.peek()) _stackMin.pop();
    }
    
    public int top() {
        if(_stack.isEmpty()) return 0;
        return (int)_stack.peek();
    }
    
    public int getMin() {
        if(_stackMin.isEmpty()) return 0;
        return (int)_stackMin.peek();
    }
}

 

posted @ 2016-09-15 05:13  咖啡中不塌缩的方糖  阅读(170)  评论(0编辑  收藏  举报