当前位置: 首页 > article >正文

算法-3-基本的数据结构

单双链表

1.单链表双链表如何反转

import java.util.ArrayList;
import java.util.List;

public class Code01_ReverseList {

	public static class Node {
		public int value;
		public Node next;

		public Node(int data) {
			value = data;
		}
	}

	public static class DoubleNode {
		public int value;
		public DoubleNode last;
		public DoubleNode next;

		public DoubleNode(int data) {
			value = data;
		}
	}

	//  head
	//   a    ->   b    ->  c  ->  null
	//   c    ->   b    ->  a  ->  null
	public static Node reverseLinkedList(Node head) {
		Node pre = null;
		Node next = null;
		while (head != null) {
			next = head.next;
			head.next = pre;
			pre = head;
			head = next;
		}
		return pre;
	}

	public static DoubleNode reverseDoubleList(DoubleNode head) {
		DoubleNode pre = null;
		DoubleNode next = null;
		while (head != null) {
			next = head.next;
			head.next = pre;
			head.last = next;
			pre = head;
			head = next;
		}
		return pre;
	}

	public static Node testReverseLinkedList(Node head) {
		if (head == null) {
			return null;
		}
		ArrayList<Node> list = new ArrayList<>();
		while (head != null) {
			list.add(head);
			head = head.next;
		}
		list.get(0).next = null;
		int N = list.size();
		for (int i = 1; i < N; i++) {
			list.get(i).next = list.get(i - 1);
		}
		return list.get(N - 1);
	}

	public static DoubleNode testReverseDoubleList(DoubleNode head) {
		if (head == null) {
			return null;
		}
		ArrayList<DoubleNode> list = new ArrayList<>();
		while (head != null) {
			list.add(head);
			head = head.next;
		}
		list.get(0).next = null;
		DoubleNode pre = list.get(0);
		int N = list.size();
		for (int i = 1; i < N; i++) {
			DoubleNode cur = list.get(i);
			cur.last = null;
			cur.next = pre;
			pre.last = cur;
			pre = cur;
		}
		return list.get(N - 1);
	}

	// for test
	public static Node generateRandomLinkedList(int len, int value) {
		int size = (int) (Math.random() * (len + 1));
		if (size == 0) {
			return null;
		}
		size--;
		Node head = new Node((int) (Math.random() * (value + 1)));
		Node pre = head;
		while (size != 0) {
			Node cur = new Node((int) (Math.random() * (value + 1)));
			pre.next = cur;
			pre = cur;
			size--;
		}
		return head;
	}

	// for test
	public static DoubleNode generateRandomDoubleList(int len, int value) {
		int size = (int) (Math.random() * (len + 1));
		if (size == 0) {
			return null;
		}
		size--;
		DoubleNode head = new DoubleNode((int) (Math.random() * (value + 1)));
		DoubleNode pre = head;
		while (size != 0) {
			DoubleNode cur = new DoubleNode((int) (Math.random() * (value + 1)));
			pre.next = cur;
			cur.last = pre;
			pre = cur;
			size--;
		}
		return head;
	}

	// for test
	public static List<Integer> getLinkedListOriginOrder(Node head) {
		List<Integer> ans = new ArrayList<>();
		while (head != null) {
			ans.add(head.value);
			head = head.next;
		}
		return ans;
	}

	// for test
	public static boolean checkLinkedListReverse(List<Integer> origin, Node head) {
		for (int i = origin.size() - 1; i >= 0; i--) {
			if (!origin.get(i).equals(head.value)) {
				return false;
			}
			head = head.next;
		}
		return true;
	}

	// for test
	public static List<Integer> getDoubleListOriginOrder(DoubleNode head) {
		List<Integer> ans = new ArrayList<>();
		while (head != null) {
			ans.add(head.value);
			head = head.next;
		}
		return ans;
	}

	// for test
	public static boolean checkDoubleListReverse(List<Integer> origin, DoubleNode head) {
		DoubleNode end = null;
		for (int i = origin.size() - 1; i >= 0; i--) {
			if (!origin.get(i).equals(head.value)) {
				return false;
			}
			end = head;
			head = head.next;
		}
		for (int i = 0; i < origin.size(); i++) {
			if (!origin.get(i).equals(end.value)) {
				return false;
			}
			end = end.last;
		}
		return true;
	}

	// for test
	public static void main(String[] args) {
		int len = 50;
		int value = 100;
		int testTime = 100000;
		System.out.println("test begin!");
		for (int i = 0; i < testTime; i++) {
			Node node1 = generateRandomLinkedList(len, value);
			List<Integer> list1 = getLinkedListOriginOrder(node1);
			node1 = reverseLinkedList(node1);
			if (!checkLinkedListReverse(list1, node1)) {
				System.out.println("Oops1!");
			}

			Node node2 = generateRandomLinkedList(len, value);
			List<Integer> list2 = getLinkedListOriginOrder(node2);
			node2 = testReverseLinkedList(node2);
			if (!checkLinkedListReverse(list2, node2)) {
				System.out.println("Oops2!");
			}

			DoubleNode node3 = generateRandomDoubleList(len, value);
			List<Integer> list3 = getDoubleListOriginOrder(node3);
			node3 = reverseDoubleList(node3);
			if (!checkDoubleListReverse(list3, node3)) {
				System.out.println("Oops3!");
			}

			DoubleNode node4 = generateRandomDoubleList(len, value);
			List<Integer> list4 = getDoubleListOriginOrder(node4);
			node4 = reverseDoubleList(node4);
			if (!checkDoubleListReverse(list4, node4)) {
				System.out.println("Oops4!");
			}

		}
		System.out.println("test finish!");

	}

}

2.把定值都删除掉

是n就跳过,不是n next指向

public class Code02_DeleteGivenValue {

	public static class Node {
		public int value;
		public Node next;

		public Node(int data) {
			this.value = data;
		}
	}

	// head = removeValue(head, 2);
	public static Node removeValue(Node head, int num) {
		// head来到第一个不需要删的位置
		while (head != null) {
			if (head.value != num) {
				break;
			}
			head = head.next;
		}
		// 1 ) head == null
		// 2 ) head != null
		Node pre = head;
		Node cur = head;
		while (cur != null) {
			if (cur.value == num) {
				pre.next = cur.next;
			} else {
				pre = cur;
			}
			cur = cur.next;
		}
		return head;
	}

}

队列跟栈

双链表实现

import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

public class Code03_DoubleEndsQueueToStackAndQueue {

	public static class Node<T> {
		public T value;
		public Node<T> last;
		public Node<T> next;

		public Node(T data) {
			value = data;
		}
	}

	public static class DoubleEndsQueue<T> {
		public Node<T> head;
		public Node<T> tail;

		public void addFromHead(T value) {
			Node<T> cur = new Node<T>(value);
			if (head == null) {
				head = cur;
				tail = cur;
			} else {
				cur.next = head;
				head.last = cur;
				head = cur;
			}
		}

		public void addFromBottom(T value) {
			Node<T> cur = new Node<T>(value);
			if (head == null) {
				head = cur;
				tail = cur;
			} else {
				cur.last = tail;
				tail.next = cur;
				tail = cur;
			}
		}

		public T popFromHead() {
			if (head == null) {
				return null;
			}
			Node<T> cur = head;
			if (head == tail) {
				head = null;
				tail = null;
			} else {
				head = head.next;
				cur.next = null;
				head.last = null;
			}
			return cur.value;
		}

		public T popFromBottom() {
			if (head == null) {
				return null;
			}
			Node<T> cur = tail;
			if (head == tail) {
				head = null;
				tail = null;
			} else {
				tail = tail.last;
				tail.next = null;
				cur.last = null;
			}
			return cur.value;
		}

		public boolean isEmpty() {
			return head == null;
		}

	}

	public static class MyStack<T> {
		private DoubleEndsQueue<T> queue;

		public MyStack() {
			queue = new DoubleEndsQueue<T>();
		}

		public void push(T value) {
			queue.addFromHead(value);
		}

		public T pop() {
			return queue.popFromHead();
		}

		public boolean isEmpty() {
			return queue.isEmpty();
		}

	}

	public static class MyQueue<T> {
		private DoubleEndsQueue<T> queue;

		public MyQueue() {
			queue = new DoubleEndsQueue<T>();
		}

		public void push(T value) {
			queue.addFromHead(value);
		}

		public T poll() {
			return queue.popFromBottom();
		}

		public boolean isEmpty() {
			return queue.isEmpty();
		}

	}

	public static boolean isEqual(Integer o1, Integer o2) {
		if (o1 == null && o2 != null) {
			return false;
		}
		if (o1 != null && o2 == null) {
			return false;
		}
		if (o1 == null && o2 == null) {
			return true;
		}
		return o1.equals(o2);
	}

	public static void main(String[] args) {
		int oneTestDataNum = 100;
		int value = 10000;
		int testTimes = 100000;
		for (int i = 0; i < testTimes; i++) {
			MyStack<Integer> myStack = new MyStack<>();
			MyQueue<Integer> myQueue = new MyQueue<>();
			Stack<Integer> stack = new Stack<>();
			Queue<Integer> queue = new LinkedList<>();
			for (int j = 0; j < oneTestDataNum; j++) {
				int nums = (int) (Math.random() * value);
				if (stack.isEmpty()) {
					myStack.push(nums);
					stack.push(nums);
				} else {
					if (Math.random() < 0.5) {
						myStack.push(nums);
						stack.push(nums);
					} else {
						if (!isEqual(myStack.pop(), stack.pop())) {
							System.out.println("oops!");
						}
					}
				}
				int numq = (int) (Math.random() * value);
				if (stack.isEmpty()) {
					myQueue.push(numq);
					queue.offer(numq);
				} else {
					if (Math.random() < 0.5) {
						myQueue.push(numq);
						queue.offer(numq);
					} else {
						if (!isEqual(myQueue.poll(), queue.poll())) {
							System.out.println("oops!");
						}
					}
				}
			}
		}
		System.out.println("finish!");
	}

}

数组实现

public class Code04_RingArray {

	public static class MyQueue {
		private int[] arr;
		private int pushi;// end
		private int polli;// begin
		private int size;
		private final int limit;

		public MyQueue(int limit) {
			arr = new int[limit];
			pushi = 0;
			polli = 0;
			size = 0;
			this.limit = limit;
		}

		public void push(int value) {
			if (size == limit) {
				throw new RuntimeException("队列满了,不能再加了");
			}
			size++;
			arr[pushi] = value;
			pushi = nextIndex(pushi);
		}

		public int pop() {
			if (size == 0) {
				throw new RuntimeException("队列空了,不能再拿了");
			}
			size--;
			int ans = arr[polli];
			polli = nextIndex(polli);
			return ans;
		}

		public boolean isEmpty() {
			return size == 0;
		}

		// 如果现在的下标是i,返回下一个位置
		private int nextIndex(int i) {
			return i < limit - 1 ? i + 1 : 0;
		}

	}

}

实现一个特殊的栈,在基本功能基础上,再实现返回栈中最小元素的功能

1.pop,push,getMin,操作的时间复杂度都是O(1)

2.设计的栈类型可以使用现成的栈结构

同步压入,同步弹出

import java.util.Stack;

public class Code05_GetMinStack {

	public static class MyStack1 {
		private Stack<Integer> stackData;
		private Stack<Integer> stackMin;

		public MyStack1() {
			this.stackData = new Stack<Integer>();
			this.stackMin = new Stack<Integer>();
		}

		public void push(int newNum) {
			if (this.stackMin.isEmpty()) {
				this.stackMin.push(newNum);
			} else if (newNum <= this.getmin()) {
				this.stackMin.push(newNum);
			}
			this.stackData.push(newNum);
		}

		public int pop() {
			if (this.stackData.isEmpty()) {
				throw new RuntimeException("Your stack is empty.");
			}
			int value = this.stackData.pop();
			if (value == this.getmin()) {
				this.stackMin.pop();
			}
			return value;
		}

		public int getmin() {
			if (this.stackMin.isEmpty()) {
				throw new RuntimeException("Your stack is empty.");
			}
			return this.stackMin.peek();
		}
	}

	public static class MyStack2 {
		private Stack<Integer> stackData;
		private Stack<Integer> stackMin;

		public MyStack2() {
			this.stackData = new Stack<Integer>();
			this.stackMin = new Stack<Integer>();
		}

		public void push(int newNum) {
			if (this.stackMin.isEmpty()) {
				this.stackMin.push(newNum);
			} else if (newNum < this.getmin()) {
				this.stackMin.push(newNum);
			} else {
				int newMin = this.stackMin.peek();
				this.stackMin.push(newMin);
			}
			this.stackData.push(newNum);
		}

		public int pop() {
			if (this.stackData.isEmpty()) {
				throw new RuntimeException("Your stack is empty.");
			}
			this.stackMin.pop();
			return this.stackData.pop();
		}

		public int getmin() {
			if (this.stackMin.isEmpty()) {
				throw new RuntimeException("Your stack is empty.");
			}
			return this.stackMin.peek();
		}
	}

	public static void main(String[] args) {
		MyStack1 stack1 = new MyStack1();
		stack1.push(3);
		System.out.println(stack1.getmin());
		stack1.push(4);
		System.out.println(stack1.getmin());
		stack1.push(1);
		System.out.println(stack1.getmin());
		System.out.println(stack1.pop());
		System.out.println(stack1.getmin());

		System.out.println("=============");

		MyStack1 stack2 = new MyStack1();
		stack2.push(3);
		System.out.println(stack2.getmin());
		stack2.push(4);
		System.out.println(stack2.getmin());
		stack2.push(1);
		System.out.println(stack2.getmin());
		System.out.println(stack2.pop());
		System.out.println(stack2.getmin());
	}

}

如何使用栈结构实现队列结构

import java.util.Stack;

public class Code06_TwoStacksImplementQueue {

	public static class TwoStacksQueue {
		public Stack<Integer> stackPush;
		public Stack<Integer> stackPop;

		public TwoStacksQueue() {
			stackPush = new Stack<Integer>();
			stackPop = new Stack<Integer>();
		}

		// push栈向pop栈倒入数据
		private void pushToPop() {
			if (stackPop.empty()) {
				while (!stackPush.empty()) {
					stackPop.push(stackPush.pop());
				}
			}
		}

		public void add(int pushInt) {
			stackPush.push(pushInt);
			pushToPop();
		}

		public int poll() {
			if (stackPop.empty() && stackPush.empty()) {
				throw new RuntimeException("Queue is empty!");
			}
			pushToPop();
			return stackPop.pop();
		}

		public int peek() {
			if (stackPop.empty() && stackPush.empty()) {
				throw new RuntimeException("Queue is empty!");
			}
			pushToPop();
			return stackPop.peek();
		}
	}

	public static void main(String[] args) {
		TwoStacksQueue test = new TwoStacksQueue();
		test.add(1);
		test.add(2);
		test.add(3);
		System.out.println(test.peek());
		System.out.println(test.poll());
		System.out.println(test.peek());
		System.out.println(test.poll());
		System.out.println(test.peek());
		System.out.println(test.poll());
	}

}

如何使用队列结构实现栈结构

import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

public class Code07_TwoQueueImplementStack {

	public static class TwoQueueStack<T> {
		public Queue<T> queue;
		public Queue<T> help;

		public TwoQueueStack() {
			queue = new LinkedList<>();
			help = new LinkedList<>();
		}

		public void push(T value) {
			queue.offer(value);
		}

		public T poll() {
			while (queue.size() > 1) {
				help.offer(queue.poll());
			}
			T ans = queue.poll();
			Queue<T> tmp = queue;
			queue = help;
			help = tmp;
			return ans;
		}

		public T peek() {
			while (queue.size() > 1) {
				help.offer(queue.poll());
			}
			T ans = queue.poll();
			help.offer(ans);
			Queue<T> tmp = queue;
			queue = help;
			help = tmp;
			return ans;
		}

		public boolean isEmpty() {
			return queue.isEmpty();
		}

	}

	public static void main(String[] args) {
		System.out.println("test begin");
		TwoQueueStack<Integer> myStack = new TwoQueueStack<>();
		Stack<Integer> test = new Stack<>();
		int testTime = 1000000;
		int max = 1000000;
		for (int i = 0; i < testTime; i++) {
			if (myStack.isEmpty()) {
				if (!test.isEmpty()) {
					System.out.println("Oops");
				}
				int num = (int) (Math.random() * max);
				myStack.push(num);
				test.push(num);
			} else {
				if (Math.random() < 0.25) {
					int num = (int) (Math.random() * max);
					myStack.push(num);
					test.push(num);
				} else if (Math.random() < 0.5) {
					if (!myStack.peek().equals(test.peek())) {
						System.out.println("Oops");
					}
				} else if (Math.random() < 0.75) {
					if (!myStack.poll().equals(test.pop())) {
						System.out.println("Oops");
					}
				} else {
					if (myStack.isEmpty() != test.isEmpty()) {
						System.out.println("Oops");
					}
				}
			}
		}

		System.out.println("test finish!");

	}

}

Master 公式求递推的时间复杂度

前提,子问题规模一致,(比如,子问题是分两个二分之N区间分别求最大值,可以Master,若分为三分之N那就不行)

公式:T(N)=a*T(N/b)+O(N的d次方)

得到abd之后

求最大值

public class Code08_GetMax {

	// 求arr中的最大值
	public static int getMax(int[] arr) {
		return process(arr, 0, arr.length - 1);
	}

	// arr[L..R]范围上求最大值  L ... R   N
	public static int process(int[] arr, int L, int R) {
		// arr[L..R]范围上只有一个数,直接返回,base case
		if (L == R) { 
			return arr[L];
		}
		// L...R 不只一个数
		// mid = (L + R) / 2
		int mid = L + ((R - L) >> 1); // 中点   	1
		int leftMax = process(arr, L, mid);
		int rightMax = process(arr, mid + 1, R);
		return Math.max(leftMax, rightMax);
	}

}

哈希

import java.util.HashMap;
import java.util.HashSet;
import java.util.TreeMap;

public class HashMapAndSortedMap {

	public static class Node {
		public int value;

		public Node(int v) {
			value = v;
		}
	}

	public static class Zuo {
		public int value;

		public Zuo(int v) {
			value = v;
		}
	}

	public static void main(String[] args) {

		HashMap<Integer, String> test = new HashMap<>();
		Integer a = 19000000;
		Integer b = 19000000;
		System.out.println(a == b);

		test.put(a, "我是3");
		System.out.println(test.containsKey(b));

		Zuo z1 = new Zuo(1);
		Zuo z2 = new Zuo(1);
		HashMap<Zuo, String> test2 = new HashMap<>();
		test2.put(z1, "我是z1");
		System.out.println(test2.containsKey(z2));

		// UnSortedMap
		HashMap<Integer, String> map = new HashMap<>();
		map.put(1000000, "我是1000000");
		map.put(2, "我是2");
		map.put(3, "我是3");
		map.put(4, "我是4");
		map.put(5, "我是5");
		map.put(6, "我是6");
		map.put(1000000, "我是1000001");

		System.out.println(map.containsKey(1));
		System.out.println(map.containsKey(10));

		System.out.println(map.get(4));
		System.out.println(map.get(10));

		map.put(4, "他是4");
		System.out.println(map.get(4));

		map.remove(4);
		System.out.println(map.get(4));

		// key
		HashSet<String> set = new HashSet<>();
		set.add("abc");
		set.contains("abc");
		set.remove("abc");

		// 哈希表,增、删、改、查,在使用时,O(1)

		System.out.println("=====================");

		Integer c = 100000;
		Integer d = 100000;
		System.out.println(c.equals(d));

		Integer e = 127; // - 128 ~ 127
		Integer f = 127;
		System.out.println(e == f);

		HashMap<Node, String> map2 = new HashMap<>();
		Node node1 = new Node(1);
		Node node2 = node1;
		map2.put(node1, "我是node1");
		map2.put(node2, "我是node1");
		System.out.println(map2.size());

		System.out.println("======================");

		// TreeMap 有序表:接口名
		// 红黑树、avl、sb树、跳表
		// O(logN)
		System.out.println("有序表测试开始");
		TreeMap<Integer, String> treeMap = new TreeMap<>();

		treeMap.put(3, "我是3");
		treeMap.put(4, "我是4");
		treeMap.put(8, "我是8");
		treeMap.put(5, "我是5");
		treeMap.put(7, "我是7");
		treeMap.put(1, "我是1");
		treeMap.put(2, "我是2");

		System.out.println(treeMap.containsKey(1));
		System.out.println(treeMap.containsKey(10));

		System.out.println(treeMap.get(4));
		System.out.println(treeMap.get(10));

		treeMap.put(4, "他是4");
		System.out.println(treeMap.get(4));

		// treeMap.remove(4);
		System.out.println(treeMap.get(4));

		System.out.println("新鲜:");

		System.out.println(treeMap.firstKey());
		System.out.println(treeMap.lastKey());
		// <= 4
		System.out.println(treeMap.floorKey(4));
		// >= 4
		System.out.println(treeMap.ceilingKey(4));
		// O(logN)

	}

}

 


http://www.kler.cn/news/234686.html

相关文章:

  • QT+OSG/osgEarth编译之八十四:osgdb_osg+Qt编译(一套代码、一套框架,跨平台编译,版本:OSG-3.6.5插件库osgdb_osg)
  • RabbitMQ——构建高性能消息传递的应用
  • 彩虹系统7.0免授权+精美WAP端模板源码
  • 基于微信小程序的校园故障维修管理系统的研究与实现
  • 探索NLP中的N-grams:理解,应用与优化
  • 【Web】Spring rce CVE-2022-22965漏洞复现学习笔记
  • 《CSS 简易速速上手小册》第8章:CSS 性能优化和可访问性(2024 最新版)
  • 格式化dingo返回内容
  • 算法竞赛进阶指南——基本算法(倍增)
  • NGINX upstream、stream、四/七层负载均衡以及案例示例
  • python从入门到精通(十八):python爬虫的练习案列集合
  • 【高阶数据结构】B-树详解
  • 如何入门AI Agent?
  • C++函数对象-运算符函数对象 - 逻辑运算 - 实现 !x 的函数对象 (std::logical_not)
  • Java 集合、迭代器
  • 跟着cherno手搓游戏引擎【24】开启2D引擎前的项目总结(包括前置知识汇总)
  • 【大厂AI课学习笔记】【1.6 人工智能基础知识】(2)机器学习
  • 07-Java桥接模式 ( Bridge Pattern )
  • 网络学习:数据链路层VLAN原理和配置
  • tkinter-TinUI-xml实战(10)展示画廊
  • mac卸载被锁定的app
  • 《CSS 简易速速上手小册》第4章:视觉美学(2024 最新版)
  • Python for 循环
  • 常见性能优化策略
  • CVE-2012-2311 漏洞复现
  • 计算机网络(第六版)复习提纲29
  • spring boot 通过 application 切换cache使用的服务
  • React18原理: 再聊Fiber架构下的时间分片
  • 前端JavaScript篇之ajax、axios、fetch的区别
  • 【LeetCode每日一题】二维前缀和基本概念与案例