leetcode 155. Min Stack 最小栈的实现

谁践踏了优雅 2022-06-08 06:30 148阅读 0赞

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.

就是使用两个栈来实现本题描述的最小栈的实现，一个栈正常使用，另一个栈保存最小值，直接上代码吧。

代码如下：

import java.util.Arrays;
    import java.util.Stack;
    
    public class MinStack 
    {
        
        private Stack<Integer> stk;
        private Stack<Integer> min;
    
        public MinStack() 
        {
            stk = new Stack<>();
            min = new Stack<>();
        }
    
        public void push(int x) 
        {
            stk.push(x);
            if(min.isEmpty() || x <= min.peek())
                min.push(x);
        }
    
        /*
         * 注意这里必须要使用equals，因为Integer是一个整数对象
         * */
        public void pop() 
        {
            if(stk.peek().equals(min.peek()))
            {
                stk.pop();
                min.pop();
            }else
                stk.pop();
        }
    
        public int top() 
        {
            return stk.peek();
        }
    
        public int getMin() 
        {
            return min.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.getMin();
     */