定義棧的數(shù)據(jù)結(jié)構(gòu)，請(qǐng)?jiān)谠擃愋椭袑?shí)現(xiàn)一個(gè)能夠得到棧中所含最小元素的min函數(shù)（時(shí)間復(fù)雜度應(yīng)為O（1））。

代碼

import java.util.Stack;
public class Solution{
    Stack<Integer> dataStack = new Stack<Integer>();
    Stack<Integer> minStack = new Stack<Integer>();
    public void push (int value){
        dataStack.push(value);
        if(minStack.isEmpty() || value<minStack.peek()){
            minStack.push(value);
        }else{
            minStack.push(minStack.peek());
        }
    }
    public void pop(){
        dataStack.pop();
        minStack.pop();
    }
    public int top(){
        return dataStack.peek();
    }
    public int min(){
        return minStack.peek();
    }

}

通過(guò)設(shè)置一個(gè)輔助棧minStack,與在dataStack中壓入得數(shù)據(jù)對(duì)比。時(shí)刻保證top的peek是最小的。