LeetCode-Min Stack

Min Stack

##题目

####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.

##解题思路
这道题实现一个最小栈的问题，这里通过压入每个元素的同时，在压入此时整个栈中的最小值。这样栈里的元素其实是压入元素与当前最小值间的交叉排列。

##算法代码
代码采用JAVA实现：

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class MinStack {
	LinkedList<Integer> stack=new LinkedList<Integer>();
    public void push(int x) {
    	if(stack.isEmpty())
    	{
    		stack.push(x);
    		stack.push(x);
    	}else{
    		int min=stack.peek();
    		stack.push(x);
	        if(x>min)
	        {
	        	stack.push(min);
	        }else{
	        	stack.push(x);
	        }
    	}
        
    }

    public void pop() {
    	if(stack.isEmpty())
    		return;
    	else
    	{
    		stack.pop();
        	stack.pop();
    	}  
    }

    public int top() {
    	if(stack.isEmpty())
    		return -1;
    	else{
    		int num=stack.pop();
	        int res=stack.peek();
	        stack.push(num);
	        return res;
    	}
       
    }

    public int getMin() {
    	if(stack.isEmpty())
    		return -1;
    	else
        	return stack.peek();
    }
}

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