README.md

30. 包含min函数的栈

定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，调用 min、push 及 pop 的时间复杂度都是 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：

各函数的调用总次数不超过 20000 次

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/bao-han-minhan-shu-de-zhan-lcof

解析

题解

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(head.min, x), 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;
        }
    }
}

/**
 * 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();
 */