leecode更新
This commit is contained in:
dingjiawen
parent
bb316cd55c
commit
806e02d386
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package com.markilue.leecode.stackAndDeque;
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import org.junit.Test;
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import java.util.*;
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/**
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* @BelongsProject: Leecode
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* @BelongsPackage: com.markilue.leecode.stackAndDeque
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* @Author: dingjiawen
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* @CreateTime: 2022-09-13 12:06
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* @Description:
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* TODO 力扣150题 逆波兰表达式求值:
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* 根据 逆波兰表示法，求表达式的值。
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* 有效的算符包括+、-、*、/。每个运算对象可以是整数，也可以是另一个逆波兰表达式。
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* 注意两个整数之间的除法只保留整数部分。
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* 可以保证给定的逆波兰表达式总是有效的。换句话说，表达式总会得出有效数值且不存在除数为 0 的情况。
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* @Version: 1.0
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*/
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public class EvalRPN {
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@Test
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public void test(){
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String[] tokens = {"2", "1", "+", "3", "*"};
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System.out.println(evalRPN(tokens));
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}
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@Test
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public void test1(){
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String[] tokens = {"4","13","5","/","+"};
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System.out.println(evalRPN(tokens));
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}
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@Test
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public void test2(){
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String[] tokens = {"10","6","9","3","+","-11","*","/","*","17","+","5","+"};
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System.out.println(evalRPN(tokens));
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}
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@Test
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public void test3(){
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String[] tokens = {"2", "1", "+", "3", "*"};
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System.out.println(evalRPN(tokens));
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}
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/**
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* 由于没有括号等操作，考虑使用一个stack解决，遇上运算符，就在往前消两栈，计算出的结果再入栈
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* 由于保证了给定的逆波兰表达式总是有效的，所以不需要额外的判断
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* 速度击败55%，内存击败47%
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* @param tokens
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* @return
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*/
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public int evalRPN(String[] tokens) {
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Stack<Integer> stack = new Stack<>();
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for (String token : tokens) {
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if(!token.equals("+")&&!token.equals("-")&&!token.equals("*")&&!token.equals("/")){
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stack.push(Integer.valueOf(token));
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}else {
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int num2=stack.pop();
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int num1=stack.pop();
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if(token.equals("+")){
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stack.push(num1+num2);
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}else if(token.equals("-")){
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stack.push(num1-num2);
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}else if(token.equals("*")){
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stack.push(num1*num2);
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}else if(token.equals("/")){
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stack.push(num1/num2);
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}
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}
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}
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return stack.pop();
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}
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/**
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* 官方给的使用数组模拟栈代码
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* @param tokens
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* @return
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*/
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public int evalRPN1(String[] tokens) {
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int n = tokens.length;
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int[] stack = new int[(n + 1) / 2];
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int index = -1;
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for (int i = 0; i < n; i++) {
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String token = tokens[i];
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switch (token) {
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case "+":
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index--;
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stack[index] += stack[index + 1];
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break;
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case "-":
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index--;
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stack[index] -= stack[index + 1];
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break;
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case "*":
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index--;
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stack[index] *= stack[index + 1];
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break;
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case "/":
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index--;
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stack[index] /= stack[index + 1];
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break;
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default:
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index++;
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stack[index] = Integer.parseInt(token);
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}
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}
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return stack[index];
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}
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}
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package com.markilue.leecode.stackAndDeque;
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import org.junit.Test;
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import java.util.Stack;
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/**
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* @BelongsProject: Leecode
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* @BelongsPackage: com.markilue.leecode.stackAndDeque
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* @Author: dingjiawen
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* @CreateTime: 2022-09-13 11:17
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* @Description: TODO 力扣20题 有效的括号:
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* 给定一个只包括 '('，')'，'{'，'}'，'['，']'的字符串 s ，判断字符串是否有效。
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* <p>
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* 有效字符串需满足：
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* 左括号必须用相同类型的右括号闭合。
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* 左括号必须以正确的顺序闭合。
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* 每个右括号都有一个对应的相同类型的左括号。
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* @Version: 1.0
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*/
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public class IsValid {
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@Test
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public void test() {
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String s = "{([([])])}";
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System.out.println(isValid(s));
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}
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/**
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* 速度击败98%，内存击败35.56%
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*
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* @param s
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* @return
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*/
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public boolean isValid(String s) {
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if (s.length() % 2 != 0) {
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return false;
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}
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Stack<Character> stack = new Stack<>();
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for (char c : s.toCharArray()) {
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if (c == '{' || c == '(' || c == '[') {
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stack.push(c);
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}
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if (c == ')') {
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if (stack.size() == 0 || stack.pop() != '(') {
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return false;
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}
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}
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if (c == '}') {
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if (stack.size() == 0 || stack.pop() != '{') {
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return false;
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}
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}
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if (c == ']') {
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if (stack.size() == 0 || stack.pop() != '[') {
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return false;
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}
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}
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}
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if (!stack.empty()) {
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return false;
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}
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return true;
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}
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/*
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某大佬巧妙思路，不过就是效率不高
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执行用时：47 ms, 在所有 Java 提交中击败了5.62%的用户
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内存消耗：41.4 MB, 在所有 Java 提交中击败了5.01%的用户
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*/
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public boolean isValid1(String s) {
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while (true) {
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int l = s.length();
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s = s.replace("()", "");
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s = s.replace("{}", "");
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s = s.replace("[]", "");
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if (s.length() == l) {
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return l == 0;
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}
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}
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}
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}
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package com.markilue.leecode.stackAndDeque;
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import org.junit.Test;
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import org.omg.CORBA.PUBLIC_MEMBER;
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import java.util.Stack;
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/**
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* @BelongsProject: Leecode
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* @BelongsPackage: com.markilue.leecode.stackAndDeque
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* @Author: dingjiawen
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* @CreateTime: 2022-09-13 09:17
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* @Description:
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* TODO 力扣232题 用栈实现队列:
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* 请你仅使用两个栈实现先入先出队列。队列应当支持一般队列支持的所有操作（push、pop、peek、empty）：
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*
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* 实现 MyQueue 类：
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* void push(int x) 将元素 x 推到队列的末尾
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* int pop() 从队列的开头移除并返回元素
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* int peek() 返回队列开头的元素
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* boolean empty() 如果队列为空，返回 true ；否则，返回 false
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*
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* 说明：
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* 你只能使用标准的栈操作 —— 也就是只有push to top,peek/pop from top,size, 和is empty操作是合法的。
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* 你所使用的语言也许不支持栈。你可以使用 list 或者 deque（双端队列）来模拟一个栈，只要是标准的栈操作即可。
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*
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* @Version: 1.0
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*/
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public class MyQueue {
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public Stack<Integer> stack1;
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public Stack<Integer> stack2;
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public MyQueue() {
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stack1=new Stack<>();
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stack2=new Stack<>();
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}
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/**
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* 本人思路:
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* stack1维护一个反向的stack,stack2就维护一个正向的stack
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* push时只push进stack1,每调用一次pop就将stack1的值push进stack2
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* 速度击败100%,内存击败50%
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* 代码随想录与本思路类似，但是他额peek复用了pop(),即pop()之后再添加回去，另外也不需要每次再pop会stack1,具体见MyQueue1
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* @param x
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*/
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public void push(int x) {
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stack1.push(x);
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}
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public int pop() {
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//放入stack2
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while (!stack1.empty()){
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stack2.push(stack1.pop());
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}
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//获取结果
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int result=stack2.pop();
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//放回stack1
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while (!stack2.empty()){
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stack1.push(stack2.pop());
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}
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//返回结果
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return result;
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}
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public int peek() {
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//放入stack2
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while (!stack1.empty()){
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stack2.push(stack1.pop());
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}
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//查看结果结果
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int result=stack2.peek();
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//放回stack1
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while (!stack2.empty()){
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stack1.push(stack2.pop());
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}
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return result;
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}
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public boolean empty() {
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return stack1.empty();
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}
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@Test
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public void test(){
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MyQueue myQueue = new MyQueue();
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myQueue.push(1); // queue is: [1]
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myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
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System.out.println(myQueue.peek());// return 1
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System.out.println(myQueue.pop());// return 1, queue is [2]
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System.out.println(myQueue.empty());// return false
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}
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}
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/**
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* Your MyQueue object will be instantiated and called as such:
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* MyQueue obj = new MyQueue();
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* obj.push(x);
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* int param_2 = obj.pop();
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* int param_3 = obj.peek();
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* boolean param_4 = obj.empty();
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*/
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package com.markilue.leecode.stackAndDeque;
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import org.junit.Test;
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import java.util.Stack;
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/**
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* @BelongsProject: Leecode
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* @BelongsPackage: com.markilue.leecode.stackAndDeque
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* @Author: dingjiawen
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* @CreateTime: 2022-09-13 09:17
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* @Description:
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* TODO 力扣232题 用栈实现队列:
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* 请你仅使用两个栈实现先入先出队列。队列应当支持一般队列支持的所有操作（push、pop、peek、empty）：
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*
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* 实现 MyQueue 类：
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* void push(int x) 将元素 x 推到队列的末尾
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* int pop() 从队列的开头移除并返回元素
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* int peek() 返回队列开头的元素
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* boolean empty() 如果队列为空，返回 true ；否则，返回 false
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*
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* 说明：
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* 你只能使用标准的栈操作 —— 也就是只有push to top,peek/pop from top,size, 和is empty操作是合法的。
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* 你所使用的语言也许不支持栈。你可以使用 list 或者 deque（双端队列）来模拟一个栈，只要是标准的栈操作即可。
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*
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* @Version: 1.0
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*/
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public class MyQueue1 {
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public Stack<Integer> stack1;
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public Stack<Integer> stack2;
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public MyQueue1() {
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stack1=new Stack<>();
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stack2=new Stack<>();
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}
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/**
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* 官方思路:
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* 但是他额peek复用了pop(),即pop()之后再添加回去，另外也不需要每次再pop会stack1,具体见MyQueue1
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* 由于充分利用了两个stack,内存利用率大大提高
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* 速度击败100%，内存击败96.07%
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* @param x
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*/
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public void push(int x) {
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stack1.push(x);
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}
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public int pop() {
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dumpStack();
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//返回结果
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return stack2.pop();
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}
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public int peek() {
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dumpStack();
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return stack2.peek();
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}
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public boolean empty() {
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return stack1.empty()&& stack2.empty();
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}
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private void dumpStack(){
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if(stack2.empty()){
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while (!stack1.empty()){
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stack2.push(stack1.pop());
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}
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}
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}
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@Test
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public void test(){
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MyQueue1 myQueue = new MyQueue1();
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myQueue.push(1); // queue is: [1]
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myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
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System.out.println(myQueue.peek());// return 1
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System.out.println(myQueue.pop());// return 1, queue is [2]
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System.out.println(myQueue.empty());// return false
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}
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}
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/**
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* Your MyQueue object will be instantiated and called as such:
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* MyQueue obj = new MyQueue();
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* obj.push(x);
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* int param_2 = obj.pop();
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* int param_3 = obj.peek();
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* boolean param_4 = obj.empty();
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*/
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@ -0,0 +1,120 @@
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package com.markilue.leecode.stackAndDeque;
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import org.junit.Test;
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import java.util.ArrayDeque;
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import java.util.Queue;
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import java.util.concurrent.PriorityBlockingQueue;
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/**
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* @BelongsProject: Leecode
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* @BelongsPackage: com.markilue.leecode.stackAndDeque
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* @Author: dingjiawen
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* @CreateTime: 2022-09-13 09:59
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* @Description: TODO 力扣225题 用队列实现栈:
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* 请你仅使用两个队列实现一个后入先出（LIFO）的栈，并支持普通栈的全部四种操作（push、top、pop 和 empty）。
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* 实现 MyStack 类：
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* void push(int x) 将元素 x 压入栈顶。
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* int pop() 移除并返回栈顶元素。
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* int top() 返回栈顶元素。
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* boolean empty() 如果栈是空的，返回 true ；否则，返回 false 。
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* <p>
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* <p>
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* 注意：
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* 你只能使用队列的基本操作 —— 也就是push to back、peek/pop from front、size 和is empty这些操作。
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* 你所使用的语言也许不支持队列。你可以使用 list （列表）或者 deque（双端队列）来模拟一个队列, 只要是标准的队列操作即可。
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* @Version: 1.0
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*/
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public class MyStack {
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public Queue<Integer> queue1;
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public Queue<Integer> queue2;
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public MyStack() {
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queue1 = new ArrayDeque<>();
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queue2 = new ArrayDeque<>();
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}
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/**
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* 上述方式是使用两个队列实现栈，事实上还可以使用一个队列实现栈,具体参考MyStack1
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* 速度击败100%，内存击败50%
|
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* @param x
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*/
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public void push(int x) {
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if(queue1.isEmpty()&&queue2.isEmpty()){
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queue1.add(x);
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return;
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}
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if(queue1.isEmpty()){
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//queue1为空,那么queue2不为空
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queue1.add(x);
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while (!queue2.isEmpty()){
|
||||
queue1.add(queue2.poll());
|
||||
}
|
||||
}else if(queue2.isEmpty()){
|
||||
//queue1为空,那么queue2为空
|
||||
queue2.add(x);
|
||||
while (!queue1.isEmpty()){
|
||||
queue2.add(queue1.poll());
|
||||
}
|
||||
}
|
||||
|
||||
|
||||
|
||||
}
|
||||
|
||||
public int pop() {
|
||||
if(!queue1.isEmpty()){
|
||||
return queue1.poll();
|
||||
}else {
|
||||
return queue2.poll();
|
||||
}
|
||||
|
||||
}
|
||||
|
||||
public int top() {
|
||||
if(!queue1.isEmpty()){
|
||||
return queue1.peek();
|
||||
}else {
|
||||
return queue2.peek();
|
||||
}
|
||||
}
|
||||
|
||||
public boolean empty() {
|
||||
return queue1.isEmpty()&&queue2.isEmpty();
|
||||
}
|
||||
|
||||
// public void dumpStack(){
|
||||
// if(queue2.isEmpty()){
|
||||
// while (!queue1.isEmpty()){
|
||||
// queue2.add(queue1.poll());
|
||||
// }
|
||||
// }
|
||||
// }
|
||||
|
||||
@Test
|
||||
public void test() {
|
||||
MyStack obj = new MyStack();
|
||||
obj.push(1);
|
||||
obj.push(2);
|
||||
obj.push(3);
|
||||
int param_2 = obj.pop();
|
||||
int param_3 = obj.top();
|
||||
boolean param_4 = obj.empty();
|
||||
|
||||
System.out.println(param_2);
|
||||
System.out.println(param_3);
|
||||
System.out.println(param_4);
|
||||
}
|
||||
}
|
||||
|
||||
/**
|
||||
* Your MyStack object will be instantiated and called as such:
|
||||
* MyStack obj = new MyStack();
|
||||
* obj.push(x);
|
||||
* int param_2 = obj.pop();
|
||||
* int param_3 = obj.top();
|
||||
* boolean param_4 = obj.empty();
|
||||
*/
|
||||
|
|
@ -0,0 +1,95 @@
|
|||
package com.markilue.leecode.stackAndDeque;
|
||||
|
||||
import org.junit.Test;
|
||||
|
||||
import java.util.ArrayDeque;
|
||||
import java.util.Queue;
|
||||
|
||||
/**
|
||||
* @BelongsProject: Leecode
|
||||
* @BelongsPackage: com.markilue.leecode.stackAndDeque
|
||||
* @Author: dingjiawen
|
||||
* @CreateTime: 2022-09-13 09:59
|
||||
* @Description: TODO 力扣225题 用队列实现栈:
|
||||
* 请你仅使用两个队列实现一个后入先出（LIFO）的栈，并支持普通栈的全部四种操作（push、top、pop 和 empty）。
|
||||
* 实现 MyStack 类：
|
||||
* void push(int x) 将元素 x 压入栈顶。
|
||||
* int pop() 移除并返回栈顶元素。
|
||||
* int top() 返回栈顶元素。
|
||||
* boolean empty() 如果栈是空的，返回 true ；否则，返回 false 。
|
||||
* <p>
|
||||
* <p>
|
||||
* 注意：
|
||||
* 你只能使用队列的基本操作 —— 也就是push to back、peek/pop from front、size 和is empty这些操作。
|
||||
* 你所使用的语言也许不支持队列。你可以使用 list （列表）或者 deque（双端队列）来模拟一个队列, 只要是标准的队列操作即可。
|
||||
* @Version: 1.0
|
||||
*/
|
||||
public class MyStack1 {
|
||||
|
||||
public Queue<Integer> queue1;
|
||||
|
||||
|
||||
public MyStack1() {
|
||||
queue1 = new ArrayDeque<>();
|
||||
}
|
||||
|
||||
/**
|
||||
* 一个队列实现stack,即先把元素加到队列尾部，然后再把头pop出在添加到尾部即可
|
||||
* 速度击败100%，内存击败17%
|
||||
*
|
||||
* @param x
|
||||
*/
|
||||
public void push(int x) {
|
||||
int size = queue1.size();
|
||||
queue1.offer(x);
|
||||
while (size-- > 0) {
|
||||
queue1.offer(queue1.poll());
|
||||
}
|
||||
|
||||
|
||||
}
|
||||
|
||||
public int pop() {
|
||||
return queue1.poll();
|
||||
}
|
||||
|
||||
public int top() {
|
||||
return queue1.peek();
|
||||
}
|
||||
|
||||
public boolean empty() {
|
||||
return queue1.isEmpty();
|
||||
}
|
||||
|
||||
// public void dumpStack(){
|
||||
// if(queue2.isEmpty()){
|
||||
// while (!queue1.isEmpty()){
|
||||
// queue2.add(queue1.poll());
|
||||
// }
|
||||
// }
|
||||
// }
|
||||
|
||||
@Test
|
||||
public void test() {
|
||||
MyStack1 obj = new MyStack1();
|
||||
obj.push(1);
|
||||
obj.push(2);
|
||||
obj.push(3);
|
||||
int param_2 = obj.pop();
|
||||
int param_3 = obj.top();
|
||||
boolean param_4 = obj.empty();
|
||||
|
||||
System.out.println(param_2);
|
||||
System.out.println(param_3);
|
||||
System.out.println(param_4);
|
||||
}
|
||||
}
|
||||
|
||||
/**
|
||||
* Your MyStack object will be instantiated and called as such:
|
||||
* MyStack obj = new MyStack();
|
||||
* obj.push(x);
|
||||
* int param_2 = obj.pop();
|
||||
* int param_3 = obj.top();
|
||||
* boolean param_4 = obj.empty();
|
||||
*/
|
||||
Loading…
Reference in New Issue