网站功能描述与分析备案期间怎么做网站

1.栈（Stack）

1.1概念

栈是一种特殊的线性表，其只允许在固定的一段进行插入和删除元素操作。进行数据插入和删除的操作的一段称为栈顶，请一段称为栈底。栈中元素遵循后进先出的原则。

压栈：栈的插入操作叫做进栈/入栈/压栈，输入的数据在栈顶。

出栈：栈的删除操作叫做出栈。

1.2栈的初始化

    public static void main(String[] args) {Stack<Integer>s=new Stack<>();s.push(1);s.push(2);s.push(3);s.push(4);}

1.3栈的模拟实现：

import java.util.Arrays;public class Mystack {private int[] elem;private int usedSize;private static final int DEFUALT_CAPACITY=10;public Mystack(int[] elem) {this.elem = new int[DEFUALT_CAPACITY];}public void push(int val){if(isFull()){elem= Arrays.copyOf(elem,2*elem.length);}elem[usedSize]=val;usedSize++;}private boolean isFull(){return usedSize==elem.length;}public int  pop(){if(isEmpty()){throw new EmptyException("栈内为空");}int oldVal=elem[usedSize-1];usedSize--;return oldVal;}private boolean isEmpty(){return usedSize==0;}public int peek(){if (isEmpty()){throw new EmptyException("栈内为空");}return elem[usedSize-1];}
}

20. 有效的括号 - 力扣（LeetCode）

class Solution {public boolean isValid(String s) {Stack<Character> stack=new Stack <>();for(int i =0;i<s.length();i++){char ch=s.charAt(i);//获取每个字符if(ch=='('||ch=='['||ch=='{'){stack.push(ch);//满足条件的再压入栈}else{if(stack.empty()){return false;//代表第一个元素是右括号无法匹配}char top=stack.peek();if((ch==')'&&top=='(')||(ch==']'&&top=='[')||(ch=='}'&&top=='{')){stack.pop();}else{return false;}//不匹配的返回}}if(!stack.empty()){return false;//最后栈中还有左括号没有被匹配}return true;}
}

150. 逆波兰表达式求值

class Solution {public int evalRPN(String[] tokens) {Stack<Integer> stack=new Stack<>();for(String s:tokens){if(isInteger(s)){stack.push(Integer.parseInt(s));}else{int num1=stack.pop();//右边int num2=stack.pop();//左边switch(s){case"+":stack.push(num1+num2);break;case"-":stack.push(-num1+num2);break;case"*":stack.push(num1*num2);break;case"/":stack.push(num2/num1);break;}}}return stack.pop();}private boolean isInteger(String s){if(s.equals("+")||s.equals("-")||s.equals("*")||s.equals("/")){return false;}return true;}
}

import java.util.Stack;public class MinStack {private Stack<Integer> stack;private Stack<Integer> minStack;public MinStack() {stack=new Stack<>();minStack=new Stack<>();}public void push(int val) {stack.push(val);if(minStack.empty()){minStack.push(val);}else {if(val<=minStack.peek()){minStack.push(val);}}}public void pop() {if(!stack.empty()){int ret=stack.pop();if(minStack.peek()==ret){minStack.pop();}}}
//获取正常栈顶元素public int top() {if(stack.empty()){return -1;}return stack.peek();}public int getMin() {if(minStack.empty()){return -1;}return minStack.peek();}
}/*** Your MinStack object will be instantiated and called as such:* MinStack obj = new MinStack();* obj.push(val);* obj.pop();* int param_3 = obj.top();* int param_4 = obj.getMin();*/

import java.util.*;public class Solution {/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可*** @param pushV int整型一维数组* @param popV int整型一维数组* @return bool布尔型*/public boolean IsPopOrder(int[] pushA,int[] popA){Stack<Integer> stack=new Stack<>();int j=0;for (int i = 0; i < pushA.length; i++) {stack.push(pushA[i]);while(!stack.empty()&&j< popA.length&&stack.peek()==popA[j]){stack.pop();j++;}}return stack.empty();}
}