Min Stack

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. 

Naive Way:正常的一个stack的话，取min需要比那里当前的stack重所有元素，当然希望4个function都是O(1)才好。那么每一次push都与当前min比较，再把这些min用指针串起来。着就是一个链表的想法，需要双向链表解决pop后找回前一个的问题。

4个函数都是O(1)，运行时间是Java里较快的。

class MinStack {
    class Node{
        int val;
        Node next;
        Node pre; 
        Node last;// pointers to next minimun Node
        Node(int x){
            val = x;
            next = null;
            pre = null;
            last = null;
        }
    }
    
    Node head = new Node(0);
    Node tail = head;
    Node min = null;
    
    public void push(int x) {
        Node node = new Node(x);
        // set stack
        tail.next = node;
        node.pre = tail;
        tail = node;
        // set min
        if(min==null){
            min = node;
        }else{
            if(node.val < min.val){
                node.last = min;
                min = node;
            }else{
                node.last = min;
            }
        }
    }

    public void pop() {
        if(tail==head){
            return;
        }else{
            min = tail.last;
            tail = tail.pre;
        }
    }

    public int top() {
        return tail.val;
    }

    public int getMin() {
        return min==null?0:min.val;
    }
}

Improved Way: Solution里说的是用两个stack,一个正常存放，一个存min，每次push之前先看看是否min和当前被push元素相等，是就同时也push 出min。这样的做法也挺好的。

class MinStack {
    Stack<Integer> stack = new Stack<Integer>();
    Stack<Integer> min = new Stack<Integer>();
    
    public void push(int x) {
        if(min.isEmpty() || (!min.isEmpty() && x <= min.peek()))
            min.push(x);
        stack.push(x);
    }

    public void pop() {
        if(!stack.isEmpty()&& !min.isEmpty()){
            if((int)stack.peek()==(int)min.peek())
                min.pop();
            stack.pop();
        }
    }

    public int top() {
        return stack.isEmpty()?0:stack.peek();
    }

    public int getMin() {
        return min.isEmpty()?0:min.peek();
    }
}



还有一个很有创意的想法是用一个stack，遇到比当前min还小的就Push一次原来的min，pop的时候遇到min时就把当前min变成stack的下一个数，然后push掉这个记号。总体就是利用原stack存当前min。来自leetcode用户sometimescrazy。

class MinStack {
    int min=Integer.MAX_VALUE;
    Stack<Integer> stack = new Stack<Integer>();
    public void push(int x) {
       // only push the old minimum value when the current 
       // minimum value changes after pushing the new value x
        if(x <= min){          
            stack.push(min);
            min=x;
        }
        stack.push(x);
    }

    public void pop() {
       // if pop operation could result in the changing of the current minimum value, 
       // pop twice and change the current minimum value to the last minimum value.
        if(stack.peek()==min) {
            stack.pop();
            min=stack.peek();
            stack.pop();
        }else{
            stack.pop();
        }
        if(stack.empty()){
            min=Integer.MAX_VALUE;
        }
    }

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

    public int getMin() {
        return min;
    }
}