算法 |《劍指offer》面試題30. 包含min函數的最小棧

定義棧的數據結構，請在該類型中實現一個能夠得到棧的最小元素的 min 函數在該棧中，調用 min、push 及 pop 的時間複雜度都是 O(1)。

示例:

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

提示：

各函數的調用總次數不超過 20000 次

題解：

class MinStack {
    private Node head;
    /** initialize your data structure here. */
    public MinStack() {

    }
    
    public void push(int x) {
        if(head == null) {
            head = new Node(x, x, null);
        }
        else {
            head = new Node(x, Math.min(x, head.min), head);
        }
    }
    
    public void pop() {
        head = head.next;
    }
    
    public int top() {
        return head.val;
    }
    
    public int min() {
        return head.min;
    }

    private class Node {
        int val;
        int min;
        Node next;

        public Node(int val, int min, Node next) {
            this.val = val;
            this.min = min;
            this.next = next;
        }
    }    
}



//雙棧法
// class MinStack {
//     Stack<Integer> s,sup;
//     /** initialize your data structure here. */
//     public MinStack() {
//         s=new Stack<Integer>();
//         sup=new Stack<Integer>();
//     }
    
//     public void push(int x) {
//         s.push(x);
//         if(sup.empty()||sup.peek()>x)sup.push(x);
//         else sup.push(sup.peek());
//     }
    
//     public void pop() {
//         sup.pop();
//         s.pop();
//     }
    
//     public int top() {
//         return s.peek();
//     }
    
//     public int min() {
//         return sup.peek();
//     }
// }


/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack obj = new MinStack();
 * obj.push(x);
 * obj.pop();
 * int param_3 = obj.top();
 * int param_4 = obj.min();
 */