常见の算法链表问题

时间复杂度

1.链表逆序

package class04;

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;
		}
	}

	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 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 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 getLinkedListOriginOrder(Node head) {
		List ans = new ArrayList<>();
		while (head != null) {
			ans.add(head.value);
			head = head.next;
		}
		return ans;
	}

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

	// for test
	public static boolean checkDoubleListReverse(List 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;
	}

	
	public static void f(Node head) {
		head = head.next;
	}
	
	// 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 list1 = getLinkedListOriginOrder(node1);
			node1 = reverseLinkedList(node1);
			if (!checkLinkedListReverse(list1, node1)) {
				System.out.println("Oops1!");
			}

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

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

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

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

	}

}

2.单链表实现堆区跟队列

package class04;

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

public class Code02_LinkedListToQueueAndStack {

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

		public Node(V v) {
			value = v;
			next = null;
		}
	}

	public static class MyQueue {
		private Node head;
		private Node tail;
		private int size;

		public MyQueue() {
			head = null;
			tail = null;
			size = 0;
		}

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

		public int size() {
			return size;
		}

		public void offer(V value) {
			Node cur = new Node(value);
			if (tail == null) {
				head = cur;
				tail = cur;
			} else {
				tail.next = cur;
				tail = cur;
			}
			size++;
		}

		// C/C++的同学需要做节点析构的工作
		public V poll() {
			V ans = null;
			if (head != null) {
				ans = head.value;
				head = head.next;
				size--;
			}
			if (head == null) {
				tail = null;
			}
			return ans;
		}

		// C/C++的同学需要做节点析构的工作
		public V peek() {
			V ans = null;
			if (head != null) {
				ans = head.value;
			}
			return ans;
		}

	}

	public static class MyStack {
		private Node head;
		private int size;

		public MyStack() {
			head = null;
			size = 0;
		}

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

		public int size() {
			return size;
		}

		public void push(V value) {
			Node cur = new Node<>(value);
			if (head == null) {
				head = cur;
			} else {
				cur.next = head;
				head = cur;
			}
			size++;
		}

		public V pop() {
			V ans = null;
			if (head != null) {
				ans = head.value;
				head = head.next;
				size--;
			}
			return ans;
		}

		public V peek() {
			return head != null ? head.value : null;
		}

	}

	public static void testQueue() {
		MyQueue myQueue = new MyQueue<>();
		Queue test = new LinkedList<>();
		int testTime = 5000000;
		int maxValue = 200000000;
		System.out.println("测试开始!");
		for (int i = 0; i < testTime; i++) {
			if (myQueue.isEmpty() != test.isEmpty()) {
				System.out.println("Oops!");
			}
			if (myQueue.size() != test.size()) {
				System.out.println("Oops!");
			}
			double decide = Math.random();
			if (decide < 0.33) {
				int num = (int) (Math.random() * maxValue);
				myQueue.offer(num);
				test.offer(num);
			} else if (decide < 0.66) {
				if (!myQueue.isEmpty()) {
					int num1 = myQueue.poll();
					int num2 = test.poll();
					if (num1 != num2) {
						System.out.println("Oops!");
					}
				}
			} else {
				if (!myQueue.isEmpty()) {
					int num1 = myQueue.peek();
					int num2 = test.peek();
					if (num1 != num2) {
						System.out.println("Oops!");
					}
				}
			}
		}
		if (myQueue.size() != test.size()) {
			System.out.println("Oops!");
		}
		while (!myQueue.isEmpty()) {
			int num1 = myQueue.poll();
			int num2 = test.poll();
			if (num1 != num2) {
				System.out.println("Oops!");
			}
		}
		System.out.println("测试结束!");
	}

	public static void testStack() {
		MyStack myStack = new MyStack<>();
		Stack test = new Stack<>();
		int testTime = 5000000;
		int maxValue = 200000000;
		System.out.println("测试开始!");
		for (int i = 0; i < testTime; i++) {
			if (myStack.isEmpty() != test.isEmpty()) {
				System.out.println("Oops!");
			}
			if (myStack.size() != test.size()) {
				System.out.println("Oops!");
			}
			double decide = Math.random();
			if (decide < 0.33) {
				int num = (int) (Math.random() * maxValue);
				myStack.push(num);
				test.push(num);
			} else if (decide < 0.66) {
				if (!myStack.isEmpty()) {
					int num1 = myStack.pop();
					int num2 = test.pop();
					if (num1 != num2) {
						System.out.println("Oops!");
					}
				}
			} else {
				if (!myStack.isEmpty()) {
					int num1 = myStack.peek();
					int num2 = test.peek();
					if (num1 != num2) {
						System.out.println("Oops!");
					}
				}
			}
		}
		if (myStack.size() != test.size()) {
			System.out.println("Oops!");
		}
		while (!myStack.isEmpty()) {
			int num1 = myStack.pop();
			int num2 = test.pop();
			if (num1 != num2) {
				System.out.println("Oops!");
			}
		}
		System.out.println("测试结束!");
	}

	public static void main(String[] args) {
		testQueue();
		testStack();
	}

}

3.双链表结构实现双端队列

package class04;

import java.util.Deque;
import java.util.LinkedList;

public class Code03_DoubleLinkedListToDeque {

	public static class Node {
		public V value;
		public Node last;
		public Node next;

		public Node(V v) {
			value = v;
			last = null;
			next = null;
		}
	}

	public static class MyDeque {
		private Node head;
		private Node tail;
		private int size;

		public MyDeque() {
			head = null;
			tail = null;
			size = 0;
		}

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

		public int size() {
			return size;
		}

		public void pushHead(V value) {	//从头部加
			Node cur = new Node<>(value);
			if (head == null) { //之前没有节点,cur是第一个
				head = cur;
				tail = cur;
			} else {
				cur.next = head;
				head.last = cur;
				head = cur;
			}
			size++;
		}

		public void pushTail(V value) {
			Node cur = new Node<>(value);
			if (head == null) {
				head = cur;
				tail = cur;
			} else {
				tail.next = cur;
				cur.last = tail;
				tail = cur;
			}
			size++;
		}

		public V pollHead() {	//从头部弹出
			V ans = null;
			if (head == null) {
				return ans;
			}
			size--;
			ans = head.value;
			if (head == tail) {	//只有一个节点
				head = null;
				tail = null;
			} else {
				head = head.next;
				head.last = null;
			}
			return ans;
		}

		public V pollTail() {
			V ans = null;
			if (head == null) {
				return ans;
			}
			size--;
			ans = tail.value;
			if (head == tail) {
				head = null;
				tail = null;
			} else {
				tail = tail.last;
				tail.next = null;
			}
			return ans;
		}

		public V peekHead() {
			V ans = null;
			if (head != null) {
				ans = head.value;
			}
			return ans;
		}

		public V peekTail() {
			V ans = null;
			if (tail != null) {
				ans = tail.value;
			}
			return ans;
		}

	}

	public static void testDeque() {
		MyDeque myDeque = new MyDeque<>();
		Deque test = new LinkedList<>();
		int testTime = 5000000;
		int maxValue = 200000000;
		System.out.println("测试开始!");
		for (int i = 0; i < testTime; i++) {
			if (myDeque.isEmpty() != test.isEmpty()) {
				System.out.println("Oops!");
			}
			if (myDeque.size() != test.size()) {
				System.out.println("Oops!");
			}
			double decide = Math.random();
			if (decide < 0.33) {
				int num = (int) (Math.random() * maxValue);
				if (Math.random() < 0.5) {
					myDeque.pushHead(num);
					test.addFirst(num);
				} else {
					myDeque.pushTail(num);
					test.addLast(num);
				}
			} else if (decide < 0.66) {
				if (!myDeque.isEmpty()) {
					int num1 = 0;
					int num2 = 0;
					if (Math.random() < 0.5) {
						num1 = myDeque.pollHead();
						num2 = test.pollFirst();
					} else {
						num1 = myDeque.pollTail();
						num2 = test.pollLast();
					}
					if (num1 != num2) {
						System.out.println("Oops!");
					}
				}
			} else {
				if (!myDeque.isEmpty()) {
					int num1 = 0;
					int num2 = 0;
					if (Math.random() < 0.5) {
						num1 = myDeque.peekHead();
						num2 = test.peekFirst();
					} else {
						num1 = myDeque.peekTail();
						num2 = test.peekLast();
					}
					if (num1 != num2) {
						System.out.println("Oops!");
					}
				}
			}
		}
		if (myDeque.size() != test.size()) {
			System.out.println("Oops!");
		}
		while (!myDeque.isEmpty()) {
			int num1 = myDeque.pollHead();
			int num2 = test.pollFirst();
			if (num1 != num2) {
				System.out.println("Oops!");
			}
		}
		System.out.println("测试结束!");
	}

	public static void main(String[] args) {
		testDeque();
	}

}

4.k个节点的组内逆序调整

常见の算法链表问题_第1张图片

reverse函数

常见の算法链表问题_第2张图片

常见の算法链表问题_第3张图片

package class04;

// 测试链接:https://leetcode.com/problems/reverse-nodes-in-k-group/
public class Code04_ReverseNodesInKGroup {

	// 不要提交这个类
	public static class ListNode {
		public int val;
		public ListNode next;
	}

	public static ListNode reverseKGroup(ListNode head, int k) {
		ListNode start = head;
		ListNode end = getKGroupEnd(start, k);
		if (end == null) {	//第一组小于k个
			return head;
		}
		// 第一组凑齐了!
		head = end;		//以后head不动了
		reverse(start, end);
		// 上一组的结尾节点
		ListNode lastEnd = start;
		while (lastEnd.next != null) {
			start = lastEnd.next;
			end = getKGroupEnd(start, k);
			if (end == null) {
				return head;
			}
			reverse(start, end);
			lastEnd.next = end;
			lastEnd = start;
		}
		return head;
	}
	//设一个函数,给你开始节点,数够k个,把第k个返回
	public static ListNode getKGroupEnd(ListNode start, int k) {
		while (--k != 0 && start != null) {		//&前面--k!=0,指针往后走,&后面是不够k个返回空
			start = start.next;				//
		}
		return start;
	}
//
	public static void reverse(ListNode start, ListNode end) {
		end = end.next;	
		ListNode pre = null;
		ListNode cur = start;
		ListNode next = null;
		while (cur != end) {	//到end才停,所以第一步end先往后跳一格
			next = cur.next;
			cur.next = pre;
			pre = cur;
			cur = next;
		}
		start.next = end;
	}

}


5.两个链表相加

常见の算法链表问题_第4张图片

package class04;

// 测试链接:https://leetcode.com/problems/add-two-numbers/
public class Code05_AddTwoNumbers {

	// 不要提交这个类
	public static class ListNode {
		public int val;
		public ListNode next;

		public ListNode(int val) {
			this.val = val;
		}

		public ListNode(int val, ListNode next) {
			this.val = val;
			this.next = next;
		}
	}

	public static ListNode addTwoNumbers(ListNode head1, ListNode head2) {
		int len1 = listLength(head1);
		int len2 = listLength(head2);
		//长链表归l,短链表归s
		ListNode l = len1 >= len2 ? head1 : head2;
		ListNode s = l == head1 ? head2 : head1;

		ListNode curL = l;
		ListNode curS = s;
		ListNode last = curL;
		int carry = 0;		//单独开一个,储存进位信息(0/1)
		int curNum = 0;
		while (curS != null) {	//第一阶段长链表短链表都有
			curNum = curL.val + curS.val + carry;	//当前位置的值
			curL.val = (curNum % 10);
			carry = curNum / 10;
			last = curL;	//last记录最后一个不空的节点(备份),直到第三阶段
			curL = curL.next;
			curS = curS.next;
		}
		while (curL != null) {	第二阶段,短链表没了
			curNum = curL.val + carry;
			curL.val = (curNum % 10);
			carry = curNum / 10;
			last = curL;
			curL = curL.next;
		}
		if (carry != 0) {	第二阶段,长短链表都没了
			last.next = new ListNode(1);
		}
		return l;	//把两个链表相加之后的结果覆盖到原来的长链表中,也可以在单独开一个链表储存相加之后的额结果
	}

	// 求链表长度
	public static int listLength(ListNode head) {
		int len = 0;
		while (head != null) {
			len++;
			head = head.next;
		}
		return len;
	}

}

6.两个有序链表的合并

常见の算法链表问题_第5张图片

package class04;

// 测试链接:https://leetcode.com/problems/merge-two-sorted-lists
public class Code06_MergeTwoSortedLinkedList {

	// 不要提交这个类
	public static class ListNode {
		public int val;
		public ListNode next;
	}

	public static ListNode mergeTwoLists(ListNode head1, ListNode head2) {
		if (head1 == null || head2 == null) {
			return head1 == null ? head2 : head1;
		}
		ListNode head = head1.val <= head2.val ? head1 : head2;	//小的做总头
		ListNode cur1 = head.next;
		ListNode cur2 = head == head1 ? head2 : head1;
		ListNode pre = head;
		while (cur1 != null && cur2 != null) {
			if (cur1.val <= cur2.val) {
				pre.next = cur1;	//pre的指向,谁小指谁
				cur1 = cur1.next;	//谁小谁往下走
			} else {
				pre.next = cur2;
				cur2 = cur2.next;
			}
			pre = pre.next;
		}
		pre.next = cur1 != null ? cur1 : cur2;
		return head;
	}

}

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