JAVA栈相关习题3

1.将递归转化为循环

比如：逆序打印链表

// 递归方式void printList(Node head){if(null != head){printList(head.next);System.out.print(head.val + " ");}}
// 循环方式void printList(Node head){if(null==head){return;}Stack<Node> s=new Stack<>();
// 将链表中的结点保存在栈中Node cur=head;while(null!=cur){s.push(cur);cur=cur.next;}// 将栈中的元素出栈while(!s.empty()){System.out.print(s.pop().val+" ");}}

public void show3(ListNode head) {if(head == null) {return;}if(head.next == null) {System.out.println(head.val);return;}show3(head.next);System.out.println(head.val);}public void show4() {Stack<ListNode> stack = new Stack<>();ListNode cur = head;while (cur != null) {stack.push(cur);cur = cur.next;}//依次出栈while (!stack.empty()) {ListNode tmp = stack.pop();System.out.println(tmp.val);}}

2.括号匹配

（1）闭合顺序不对([)]

（2）左括号多了(()

（3）右括号多了())

结论：

1.当括号是匹配的时候，最终i遍历完了字符串，并且栈为空

2.遇到不匹配的就直接结束

3.当i遍历完了字符串，但是栈当中仍然有括号，则必定是左括号多了（肯定不匹配）

4.当栈空了，但是字符串还没有遍历完，则必定不匹配

思考：如何判断匹配

https://leetcode.cn/problems/valid-parentheses/description/

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);if(ch=='('||ch=='['||ch=='{'){//左括号入栈，因为左括号一定是符号最开始的部分stack.push(ch);}else{//是右括号的情况if(stack.empty()){return false;}else{char tmp=stack.peek();if((tmp=='('&&ch==')')||(tmp=='['&&ch==']')||(tmp=='{'&&ch=='}')){//括号匹配成功stack.pop();}else{return false;}}}}if(!stack.empty()){return false;}return true;}
}

3.逆波兰表达式

150. 逆波兰表达式求值 - 力扣（LeetCode）

class Solution {public int evalRPN(String[] tokens) {Stack<Integer> stack=new Stack<>();for(String s:tokens){if(!isOpera(s)){//数字：放入栈当中stack.push(Integer.parseInt(s));//将字符串转成整数}else{//弹出栈顶的两个元素int num2=stack.pop();int num1=stack.pop();switch(s){case "+":stack.push(num1+num2);break;case "-":stack.push(num1-num2);break;case "*":stack.push(num1*num2);break;case "/":stack.push(num1/num2);break;}}}return stack.pop();}public boolean isOpera(String s){if(s.equals("+")||s.equals("-")||s.equals("*")||s.equals("/")){return true;}return false;}
}

4.栈的压入、弹出序列

1）何时入栈---每次都入

2）什么时候出栈

栈不为空的时候并且栈顶元素和k下标的元素相同时可以出栈

栈的压入、弹出序列_牛客题霸_牛客网 (nowcoder.com)

import java.util.*;public class Solution {public boolean IsPopOrder(int [] pushA,int [] popA) {Stack<Integer> stack = new Stack<>();int j = 0;//变量popA这个数组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();}
}

5.最小栈

import java.util.Stack;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);//第一次在最小栈当中存储元素if(minStack.empty()) {minStack.push(val);}else {if(val <= minStack.peek()) {minStack.push(val);}}}public void pop() {//栈为空 则不能进行弹出元素if(stack.empty()) {return;}int val = stack.pop();if(val == minStack.peek()) {minStack.pop();}}//获取栈顶元素 和 最小栈没有关系public int top() {if(stack.empty()) {return -1;}return stack.peek();}//获取元素 不是删除元素public int getMin() {return minStack.peek();}}