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博客目录
- 一.栈介绍
- 1.概述
- 2.链表实现
- 3.数组实现
- 4.应用
- 二.栈题目
- 1.有效的括号-力扣 20 题
- 2.后缀表达式求值-力扣 120 题
- 3.中缀表达式转后缀
- 4.双栈模拟队列-力扣 232 题
- 5.单队列模拟栈-力扣 225 题
一.栈介绍
1.概述
计算机科学中，stack 是一种线性的数据结构，只能在其一端添加数据和移除数据。习惯来说，这一端称之为栈顶，另一端不能操作数据的称之为栈底，就如同生活中的一摞书
先提供一个栈接口
public interface Stack<E> {/*** 向栈顶压入元素* @param value 待压入值* @return 压入成功返回 true, 否则返回 false*/boolean push(E value);/*** 从栈顶弹出元素* @return 栈非空返回栈顶元素, 栈为空返回 null*/E pop();/*** 返回栈顶元素, 不弹出* @return 栈非空返回栈顶元素, 栈为空返回 null*/E peek();/*** 判断栈是否为空* @return 空返回 true, 否则返回 false*/boolean isEmpty();/*** 判断栈是否已满* @return 满返回 true, 否则返回 false*/boolean isFull();
}
2.链表实现
public class LinkedListStack<E> implements Stack<E>, Iterable<E> {private final int capacity;private int size;private final Node<E> head = new Node<>(null, null);public LinkedListStack(int capacity) {this.capacity = capacity;}@Overridepublic boolean push(E value) {if (isFull()) {return false;}head.next = new Node<>(value, head.next);size++;return true;}@Overridepublic E pop() {if (isEmpty()) {return null;}Node<E> first = head.next;head.next = first.next;size--;return first.value;}@Overridepublic E peek() {if (isEmpty()) {return null;}return head.next.value;}@Overridepublic boolean isEmpty() {return head.next == null;}@Overridepublic boolean isFull() {return size == capacity;}@Overridepublic Iterator<E> iterator() {return new Iterator<E>() {Node<E> p = head.next;@Overridepublic boolean hasNext() {return p != null;}@Overridepublic E next() {E value = p.value;p = p.next;return value;}};}static class Node<E> {E value;Node<E> next;public Node(E value, Node<E> next) {this.value = value;this.next = next;}}
}
3.数组实现
public class ArrayStack<E> implements Stack<E>, Iterable<E>{private final E[] array;private int top = 0;@SuppressWarnings("all")public ArrayStack(int capacity) {this.array = (E[]) new Object[capacity];}@Overridepublic boolean push(E value) {if (isFull()) {return false;}array[top++] = value;return true;}@Overridepublic E pop() {if (isEmpty()) {return null;}return array[--top];}@Overridepublic E peek() {if (isEmpty()) {return null;}return array[top-1];}@Overridepublic boolean isEmpty() {return top == 0;}@Overridepublic boolean isFull() {return top == array.length;}@Overridepublic Iterator<E> iterator() {return new Iterator<E>() {int p = top;@Overridepublic boolean hasNext() {return p > 0;}@Overridepublic E next() {return array[--p];}};}
}
4.应用
模拟如下方法调用
public static void main(String[] args) {System.out.println("main1");System.out.println("main2");method1();method2();System.out.println("main3");
}public static void method1() {System.out.println("method1");method3();
}public static void method2() {System.out.println("method2");
}public static void method3() {System.out.println("method3");
}
模拟代码
public class CPU {static class Frame {int exit;public Frame(int exit) {this.exit = exit;}}static int pc = 1; // 模拟程序计数器 Program counterstatic ArrayStack<Frame> stack = new ArrayStack<>(100); // 模拟方法调用栈public static void main(String[] args) {stack.push(new Frame(-1));while (!stack.isEmpty()) {switch (pc) {case 1 -> {System.out.println("main1");pc++;}case 2 -> {System.out.println("main2");pc++;}case 3 -> {stack.push(new Frame(pc + 1));pc = 100;}case 4 -> {stack.push(new Frame(pc + 1));pc = 200;}case 5 -> {System.out.println("main3");pc = stack.pop().exit;}case 100 -> {System.out.println("method1");stack.push(new Frame(pc + 1));pc = 300;}case 101 -> {pc = stack.pop().exit;}case 200 -> {System.out.println("method2");pc = stack.pop().exit;}case 300 -> {System.out.println("method3");pc = stack.pop().exit;}}}}
}
二.栈题目
1.有效的括号-力扣 20 题
一个字符串中可能出现 [] () 和 {} 三种括号，判断该括号是否有效
有效的例子
()[]{}([{}])()
无效的例子
[)([)]([]
思路
- 遇到左括号, 把要配对的右括号放入栈顶
- 遇到右括号, 若此时栈为空, 返回 false，否则把它与栈顶元素对比
- 若相等, 栈顶元素弹出, 继续对比下一组
- 若不等, 无效括号直接返回 false
- 循环结束
- 若栈为空, 表示所有括号都配上对, 返回 true
- 若栈不为空, 表示右没配对的括号, 应返回 false
答案（用到了课堂案例中的 ArrayStack 类）
public boolean isValid(String s) {ArrayStack<Character> stack = new ArrayStack<>(s.length() / 2 + 1);for (int i = 0; i < s.length(); i++) {char c = s.charAt(i);if (c == '(') {stack.push(')');} else if (c == '[') {stack.push(']');} else if (c == '{') {stack.push('}');} else {if (!stack.isEmpty() && stack.peek() == c) {stack.pop();} else {return false;}}}return stack.isEmpty();
}
2.后缀表达式求值-力扣 120 题
后缀表达式也称为逆波兰表达式，即运算符写在后面
- 从左向右进行计算
- 不必考虑运算符优先级，即不用包含括号
示例
输入：tokens = ["2","1","+","3","*"]
输出：9
即：(2 + 1) * 3输入：tokens = ["4","13","5","/","+"]
输出：6
即：4 + (13 / 5)
题目假设
- 数字都视为整数
- 数字和运算符个数给定正确，不会有除零发生
代码
public int evalRPN(String[] tokens) {LinkedList<Integer> numbers = new LinkedList<>();for (String t : tokens) {switch (t) {case "+" -> {Integer b = numbers.pop();Integer a = numbers.pop();numbers.push(a + b);}case "-" -> {Integer b = numbers.pop();Integer a = numbers.pop();numbers.push(a - b);}case "*" -> {Integer b = numbers.pop();Integer a = numbers.pop();numbers.push(a * b);}case "/" -> {Integer b = numbers.pop();Integer a = numbers.pop();numbers.push(a / b);}default -> numbers.push(Integer.parseInt(t));}}return numbers.pop();
}
3.中缀表达式转后缀
#反编译可以看到执行指令的过程
javap -c -v .\E03InfixToSuffix.class
public class E03InfixToSuffix {/*思路1. 遇到数字, 拼串2. 遇到 + - * /- 优先级高于栈顶运算符 入栈- 否则将栈中高级或平级运算符出栈拼串, 本运算符入栈3. 遍历完成, 栈中剩余运算符出栈拼串- 先出栈,意味着优先运算4. 带 ()- 左括号直接入栈- 右括号要将栈中直至左括号为止的运算符出栈拼串| || || |_____a+ba+b-ca+b*ca*b+c(a+b)*c*/public static void main(String[] args) {System.out.println(infixToSuffix("a+b"));System.out.println(infixToSuffix("a+b-c"));System.out.println(infixToSuffix("a+b*c"));System.out.println(infixToSuffix("a*b-c"));System.out.println(infixToSuffix("(a+b)*c"));System.out.println(infixToSuffix("a+b*c+(d*e+f)*g"));}static String infixToSuffix(String exp) {LinkedList<Character> stack = new LinkedList<>();StringBuilder sb = new StringBuilder(exp.length());for (int i = 0; i < exp.length(); i++) {char c = exp.charAt(i);switch (c) {case '+', '-', '*', '/' -> {if (stack.isEmpty()) {stack.push(c);} else {if (priority(c) > priority(stack.peek())) {stack.push(c);} else {while (!stack.isEmpty()&& priority(stack.peek()) >= priority(c)) {sb.append(stack.pop());}stack.push(c);}}}case '(' -> {stack.push(c);}case ')' -> {while (!stack.isEmpty() && stack.peek() != '(') {sb.append(stack.pop());}stack.pop();}default -> {sb.append(c);}}}while (!stack.isEmpty()) {sb.append(stack.pop());}return sb.toString();}static int priority(char c) {return switch (c) {case '(' -> 0;case '*', '/' -> 2;case '+', '-' -> 1;default -> throw new IllegalArgumentException("不合法字符:" + c);};}
}
4.双栈模拟队列-力扣 232 题
给力扣题目用的自实现栈，可以定义为静态内部类
class ArrayStack<E> {private E[] array;private int top; // 栈顶指针@SuppressWarnings("all")public ArrayStack(int capacity) {this.array = (E[]) new Object[capacity];}public boolean push(E value) {if (isFull()) {return false;}array[top++] = value;return true;}public E pop() {if (isEmpty()) {return null;}return array[--top];}public E peek() {if (isEmpty()) {return null;}return array[top - 1];}public boolean isEmpty() {return top == 0;}public boolean isFull() {return top == array.length;}
}
参考解答，注意：题目已说明
- 调用 push、pop 等方法的次数最多 100
public class E04Leetcode232 {/*队列头 队列尾s1 s2顶 底 底 顶abcpush(a)push(b)push(c)pop()*/ArrayStack<Integer> s1 = new ArrayStack<>(100);ArrayStack<Integer> s2 = new ArrayStack<>(100);public void push(int x) {s2.push(x);}public int pop() {if (s1.isEmpty()) {while (!s2.isEmpty()) {s1.push(s2.pop());}}return s1.pop();}public int peek() {if (s1.isEmpty()) {while (!s2.isEmpty()) {s1.push(s2.pop());}}return s1.peek();}public boolean empty() {return s1.isEmpty() && s2.isEmpty();}}
5.单队列模拟栈-力扣 225 题
给力扣题目用的自实现队列，可以定义为静态内部类
public class ArrayQueue3<E> {private final E[] array;int head = 0;int tail = 0;@SuppressWarnings("all")public ArrayQueue3(int c) {c -= 1;c |= c >> 1;c |= c >> 2;c |= c >> 4;c |= c >> 8;c |= c >> 16;c += 1;array = (E[]) new Object[c];}public boolean offer(E value) {if (isFull()) {return false;}array[tail & (array.length - 1)] = value;tail++;return true;}public E poll() {if (isEmpty()) {return null;}E value = array[head & (array.length - 1)];head++;return value;}public E peek() {if (isEmpty()) {return null;}return array[head & (array.length - 1)];}public boolean isEmpty() {return head == tail;}public boolean isFull() {return tail - head == array.length;}
}
参考解答，注意：题目已说明
- 调用 push、pop 等方法的次数最多 100
- 每次调用 pop 和 top 都能保证栈不为空
public class E05Leetcode225 {/*队列头 队列尾cba顶 底queue.offer(a)queue.offer(b)queue.offer(c)*/ArrayQueue3<Integer> queue = new ArrayQueue3<>(100);int size = 0;public void push(int x) {queue.offer(x);for (int i = 0; i < size; i++) {queue.offer(queue.poll());}size++;}public int pop() {size--;return queue.poll();}public int top() {return queue.peek();}public boolean empty() {return queue.isEmpty();}
}
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