尚硅谷04 数据结构与算法

文章目录

单链表面试题
1. 单链表中有效节点的个数
- 1.1 思路
- 1.2 代码实现
2. 查找单链表中的倒数第k个节点
- 2.1 思路
- 2.2 代码实现
3. 单链表的反转
- 3.1 思路
- 3.2 代码实现
4. 从尾到头打印单链表
- 4.1 思路
- 4.2 代码实现
5. 合并两个有序的单链表，合并后依然有序
- 5.1 思路
- 5.2 实现

单链表面试题

以下实现的方法都是写在SingleLinkedListDemo类里的一个静态方法

1. 单链表中有效节点的个数

1.1 思路

循环遍历单链表，用一个变量 length 记录节点个数，节点不为空 length+1

1.2 代码实现

package com.SingleLinkedListPrac04;

public class SingleLinkedListDemo {
    
    public static void main(String[] args) {
        HeroNode heroNode1 = new HeroNode(1, "宋江", "及时雨");
        HeroNode heroNode2 = new HeroNode(2, "晁盖", "玉麒麟");
        HeroNode heroNode3 = new HeroNode(3, "吴用", "智多星");

        SingleLinkedList singleLinkedList = new SingleLinkedList();
        singleLinkedList.add(heroNode1);
        singleLinkedList.add(heroNode2);
        singleLinkedList.add(heroNode3);
        singleLinkedList.list();

        //有效节点的个数测试
        System.out.println("有效节点的个数为：" + getLength(singleLinkedList.getHead()));

    }

    // 9. 单链表中有效节点的个数
    public static int getLength(HeroNode head) {
        int length = 0;
        HeroNode temp = head;
        while (temp.next != null) {
            length++;
            temp = temp.next;
        }
        return length;
    }
    // 10. 查找单链表中的倒数第k个节点
    // 11. 单链表的反转
    // 12. 从尾到头打印单链表
    // 13. 合并两个有序的单链表，合并后依然有序
}

class HeroNode {
    // 1. 创建一个节点类，用来存放节点
    // 2. 添加节点类的属性
    public int no;
    public String name;
    public String nickname;
    public HeroNode next;
    // 3. 添加构造方法，初始化属性
    public HeroNode(int no, String name, String nickname) {
        this.no = no;
        this.name = name;
        this.nickname = nickname;
    }
    // 4. toString方法，显示节点信息

    @Override
    public String toString() {
        return "HeroNode{" +
                "no=" + no +
                ", name='" + name + ''' +
                ", nickname='" + nickname + ''' +
                '}';
    }
}

class SingleLinkedList {
    // 5. 创建一个单链表的类，用来添加对单链表的操作
    // 6. 添加节点
    private HeroNode head = new HeroNode(0, "", "");
    public HeroNode getHead() {
        return this.head;
    }
    public void add(HeroNode heroNode) {
        HeroNode temp = head;
        while (temp.next != null) {
            temp = temp.next;
        }
        temp.next = heroNode;
    }
    // 7. 有序添加节点
    public void addByOrder(HeroNode heroNode) {
        HeroNode temp = head;
        while (temp.next != null) {
            if (temp.next.no < heroNode.no) {
                temp = temp.next;
            }else if (temp.next.no > heroNode.no) {
                heroNode.next = temp.next;
                break;
            }else {
                System.out.printf("编号为 %d 的节点已经存在，添加失败~~~n", heroNode.no);
                return;
            }
            temp.next = heroNode;
        }
    }
    // 8. 显示单链表的节点信息
    public void list() {
        if (head.next == null) {
            System.out.println("该单链表为空，没有可以显示的节点信息~~~");
            return;
        }
        HeroNode temp = head;
        while (temp.next != null) {
            System.out.println(temp.next);
            temp = temp.next;
        }
    }
}

2. 查找单链表中的倒数第k个节点

2.1 思路

单链表的的查找我们一般是顺序查找的，这里要求倒着查找第 k 个节点
可以先获取单链表的长度 length 后，顺序查找 length-k 个节点

2.2 代码实现

package com.SingleLinkedListPrac04;

public class SingleLinkedListDemo {
    
    public static void main(String[] args) {
        HeroNode heroNode1 = new HeroNode(1, "宋江", "及时雨");
        HeroNode heroNode2 = new HeroNode(2, "晁盖", "玉麒麟");
        HeroNode heroNode3 = new HeroNode(3, "吴用", "智多星");

        SingleLinkedList singleLinkedList = new SingleLinkedList();
        singleLinkedList.add(heroNode1);
        singleLinkedList.add(heroNode2);
        singleLinkedList.add(heroNode3);
        singleLinkedList.list();

        //有效节点的个数测试
        System.out.println("有效节点的个数为：" + getLength(singleLinkedList.getHead()));

        //单链表中的倒数第k个节点测试
        HeroNode res = findLastIndexNode(singleLinkedList.getHead(), 1);
        System.out.println("res = " + res);

    }

    // 9. 单链表中有效节点的个数
    public static int getLength(HeroNode head) {
        if (head.next == null) {
            return 0;
        }
        int length = 0;
        HeroNode temp = head;
        while (temp.next != null) {
            length++;
            temp = temp.next;
        }
        return length;
    }
    // 10. 查找单链表中的倒数第k个节点
    public static HeroNode findLastIndexNode(HeroNode head, int k) {
        if (head.next == null) {
            return null;
        }
        int length = getLength(head);
        if (k <= 0 || k > length) {
            return null;
        }
        HeroNode cur = head.next;
        for (int i = 0; i < (length - k); i++) {
            cur = cur.next;
        }
        return cur;
    }
    // 11. 单链表的反转
    // 12. 从尾到头打印单链表
    // 13. 合并两个有序的单链表，合并后依然有序
}

class HeroNode {
    // 1. 创建一个节点类，用来存放节点
    // 2. 添加节点类的属性
    public int no;
    public String name;
    public String nickname;
    public HeroNode next;
    // 3. 添加构造方法，初始化属性
    public HeroNode(int no, String name, String nickname) {
        this.no = no;
        this.name = name;
        this.nickname = nickname;
    }
    // 4. toString方法，显示节点信息

    @Override
    public String toString() {
        return "HeroNode{" +
                "no=" + no +
                ", name='" + name + ''' +
                ", nickname='" + nickname + ''' +
                '}';
    }
}

class SingleLinkedList {
    // 5. 创建一个单链表的类，用来添加对单链表的操作
    // 6. 添加节点
    private HeroNode head = new HeroNode(0, "", "");
    public HeroNode getHead() {
        return this.head;
    }
    public void add(HeroNode heroNode) {
        HeroNode temp = head;
        while (temp.next != null) {
            temp = temp.next;
        }
        temp.next = heroNode;
    }
    // 7. 有序添加节点
    public void addByOrder(HeroNode heroNode) {
        HeroNode temp = head;
        while (temp.next != null) {
            if (temp.next.no < heroNode.no) {
                temp = temp.next;
            }else if (temp.next.no > heroNode.no) {
                heroNode.next = temp.next;
                break;
            }else {
                System.out.printf("编号为 %d 的节点已经存在，添加失败~~~n", heroNode.no);
                return;
            }
            temp.next = heroNode;
        }
    }
    // 8. 显示单链表的节点信息
    public void list() {
        if (head.next == null) {
            System.out.println("该单链表为空，没有可以显示的节点信息~~~");
            return;
        }
        HeroNode temp = head;
        while (temp.next != null) {
            System.out.println(temp.next);
            temp = temp.next;
        }
    }
}

3. 单链表的反转

3.1 思路

方法一：

利用头插法，将原链表的数据添加到新链表中（双指针）
然后让原链表的表头指向新链表的数据

方法二：

也可以直接用头插法，将数据添加到原链表的表头上。

缺点：会改变原链表的结构

3.2 代码实现 4. 从尾到头打印单链表

演示栈的基本使用：

4.1 思路

方法一：

将单链表反转后，再打印输出。由于会改变单链表的结构，所以不推荐使用

方法二：

利用栈的先进后出特性。将节点存放到栈中后，再取出。

4.2 代码实现

package com.SingleLinkedListPrac04;

import java.util.Stack;

public class SingleLinkedListDemo {
    
    public static void main(String[] args) {
        HeroNode heroNode1 = new HeroNode(1, "宋江", "及时雨");
        HeroNode heroNode2 = new HeroNode(2, "晁盖", "玉麒麟");
        HeroNode heroNode3 = new HeroNode(3, "吴用", "智多星");

        SingleLinkedList singleLinkedList = new SingleLinkedList();
        singleLinkedList.add(heroNode1);
        singleLinkedList.add(heroNode2);
        singleLinkedList.add(heroNode3);
        singleLinkedList.list();

        //有效节点的个数测试
        System.out.println("有效节点的个数为：" + getLength(singleLinkedList.getHead()));

        //单链表中的倒数第k个节点测试
        HeroNode res = findLastIndexNode(singleLinkedList.getHead(), 1);
        System.out.println("res = " + res);

        //从尾到头打印列表测试
        System.out.println("从尾到头打印列表：");
        reversePrint(singleLinkedList.getHead());
    }

    // 9. 单链表中有效节点的个数
    public static int getLength(HeroNode head) {
        if (head.next == null) {
            return 0;
        }
        int length = 0;
        HeroNode temp = head;
        while (temp.next != null) {
            length++;
            temp = temp.next;
        }
        return length;
    }
    // 10. 查找单链表中的倒数第k个节点
    public static HeroNode findLastIndexNode(HeroNode head, int k) {
        if (head.next == null) {
            return null;
        }
        int length = getLength(head);
        if (k <= 0 || k > length) {
            return null;
        }
        HeroNode cur = head.next;
        for (int i = 0; i < (length - k); i++) {
            cur = cur.next;
        }
        return cur;
    }
    // 11. 单链表的反转
    public static void reverseList(HeroNode head) {
        if (head.next == null || head.next.next == null) {
            return;
        }
        // 反转后的新链表的表头
        HeroNode reverseHead = new HeroNode(0, "", "");
        HeroNode cur = head.next;
        HeroNode next = null; //用来指向当前节点的下一个节点
        while (cur != null) {
            next = cur.next;
            cur.next = reverseHead.next;
            reverseHead.next = cur;
            cur = next;
        }
        head.next = reverseHead.next;
    }
    // 12. 从尾到头打印单链表
    public static void reversePrint(HeroNode head) {
        if (head.next == null) {
            return; // 空列表，不能打印
        }
        Stack stack = new Stack<>();
        HeroNode cur = head.next;
        while (cur != null) {
            stack.push(cur);
            cur = cur.next;
        }
        while (stack.size() > 0) {
            System.out.println(stack.pop());
        }
    }
    // 13. 合并两个有序的单链表，合并后依然有序
}

class HeroNode {
    // 1. 创建一个节点类，用来存放节点
    // 2. 添加节点类的属性
    public int no;
    public String name;
    public String nickname;
    public HeroNode next;
    // 3. 添加构造方法，初始化属性
    public HeroNode(int no, String name, String nickname) {
        this.no = no;
        this.name = name;
        this.nickname = nickname;
    }
    // 4. toString方法，显示节点信息

    @Override
    public String toString() {
        return "HeroNode{" +
                "no=" + no +
                ", name='" + name + ''' +
                ", nickname='" + nickname + ''' +
                '}';
    }
}

class SingleLinkedList {
    // 5. 创建一个单链表的类，用来添加对单链表的操作
    // 6. 添加节点
    private HeroNode head = new HeroNode(0, "", "");
    public HeroNode getHead() {
        return this.head;
    }
    public void add(HeroNode heroNode) {
        HeroNode temp = head;
        while (temp.next != null) {
            temp = temp.next;
        }
        temp.next = heroNode;
    }
    // 7. 有序添加节点
    public void addByOrder(HeroNode heroNode) {
        HeroNode temp = head;
        while (temp.next != null) {
            if (temp.next.no < heroNode.no) {
                temp = temp.next;
            }else if (temp.next.no > heroNode.no) {
                heroNode.next = temp.next;
                break;
            }else {
                System.out.printf("编号为 %d 的节点已经存在，添加失败~~~n", heroNode.no);
                return;
            }
            temp.next = heroNode;
        }
    }
    // 8. 显示单链表的节点信息
    public void list() {
        if (head.next == null) {
            System.out.println("该单链表为空，没有可以显示的节点信息~~~");
            return;
        }
        HeroNode temp = head;
        while (temp.next != null) {
            System.out.println(temp.next);
            temp = temp.next;
        }
    }
}

5. 合并两个有序的单链表，合并后依然有序

5.1 思路

如果有一个单链表为空，则直接合并另一个单链表即可
当两个单链表都不为空时，比较两个单链表的no，通过三个临时变量（cur、cur1、cur2）来控制链表数据的移动。cur控制合并后的链表、cur1控制第一个链表、cur2控制第二个链表的数据移动
直到其中一个单链表为空时，把另一个链表直接添加进来，最后返回单链表对象即可

5.2 实现

package com.SingleLinkedListPrac04;

import java.util.Stack;

public class SingleLinkedListDemo {
    
    public static void main(String[] args) {
        HeroNode heroNode1 = new HeroNode(1, "宋江", "及时雨");
        HeroNode heroNode5 = new HeroNode(5, "公孙胜", "入云龙");
        HeroNode heroNode6 = new HeroNode(6, "关胜", "大刀");


        SingleLinkedList singleLinkedList = new SingleLinkedList();
        singleLinkedList.add(heroNode1);
        singleLinkedList.add(heroNode5);
        singleLinkedList.add(heroNode6);


        System.out.println("第一个有序列表：");
        singleLinkedList.list();

        //合并两个有序的单链表，合并后依然有序测试
        HeroNode heroNode2 = new HeroNode(2, "晁盖", "玉麒麟");
        HeroNode heroNode3 = new HeroNode(3, "吴用", "智多星");
        HeroNode heroNode4 = new HeroNode(4, "林冲", "豹子头");
        SingleLinkedList singleLinkedList2 = new SingleLinkedList();
        singleLinkedList2.add(heroNode2);
        singleLinkedList2.add(heroNode3);
        singleLinkedList2.add(heroNode4);
        System.out.println("第二个有序列表：");
        singleLinkedList2.list();

        System.out.println("合并后的有序列表：");
        SingleLinkedList mergedSingleLinkedList = mergeOrderedLink(singleLinkedList.getHead(), singleLinkedList2.getHead());
        mergedSingleLinkedList.list();
    }

    // 9. 单链表中有效节点的个数
    public static int getLength(HeroNode head) {
        if (head.next == null) {
            return 0;
        }
        int length = 0;
        HeroNode temp = head;
        while (temp.next != null) {
            length++;
            temp = temp.next;
        }
        return length;
    }
    // 10. 查找单链表中的倒数第k个节点
    public static HeroNode findLastIndexNode(HeroNode head, int k) {
        if (head.next == null) {
            return null;
        }
        int length = getLength(head);
        if (k <= 0 || k > length) {
            return null;
        }
        HeroNode cur = head.next;
        for (int i = 0; i < (length - k); i++) {
            cur = cur.next;
        }
        return cur;
    }
    // 11. 单链表的反转
    public static void reverseList(HeroNode head) {
        if (head.next == null || head.next.next == null) {
            return;
        }
        // 反转后的新链表的表头
        HeroNode reverseHead = new HeroNode(0, "", "");
        HeroNode cur = head.next;
        HeroNode next = null; //用来指向当前节点的下一个节点
        while (cur != null) {
            next = cur.next;
            cur.next = reverseHead.next;
            reverseHead.next = cur;
            cur = next;
        }
        head.next = reverseHead.next;
    }
    // 12. 从尾到头打印单链表
    public static void reversePrint(HeroNode head) {
        if (head.next == null) {
            return; // 空列表，不能打印
        }
        Stack stack = new Stack<>();
        HeroNode cur = head.next;
        while (cur != null) {
            stack.push(cur);
            cur = cur.next;
        }
        while (stack.size() > 0) {
            System.out.println(stack.pop());
        }
    }
    // 13. 合并两个有序的单链表，合并后依然有序
    
    public static SingleLinkedList mergeOrderedLink(HeroNode head1, HeroNode head2) {
        HeroNode mergedHeroNode = new HeroNode(0, "", "");
        SingleLinkedList mergedSingleLinkedList = new SingleLinkedList();
        if (head1.next == null) {
            mergedSingleLinkedList.add(head2.next);
            return mergedSingleLinkedList;
        }
        if (head2.next == null) {
            mergedSingleLinkedList.add(head1.next);
            return mergedSingleLinkedList;
        }
        HeroNode cur = mergedHeroNode;
        HeroNode cur1 = head1.next;
        HeroNode cur2 = head2.next;
        while (cur1 != null && cur2 != null) {
            if (cur1.no < cur2.no) {
                cur.next = cur1;
                cur = cur.next;
                cur1 = cur1.next;
            }else if (cur1.no > cur2.no) {
                cur.next = cur2;
                cur = cur.next;
                cur2 = cur2.next;
            }else {
                mergedSingleLinkedList.add(cur1);
                cur.next = cur1.next;
                cur = cur.next;
                cur1 = cur1.next;
                cur2 = cur2.next;
            }
        }
        if (cur1 == null) {
            cur.next = cur2;
        }else if (cur2 == null) {
            cur.next = cur1;
        }
        mergedSingleLinkedList.add(mergedHeroNode.next);
        return mergedSingleLinkedList;
    }
}

class HeroNode {
    // 1. 创建一个节点类，用来存放节点
    // 2. 添加节点类的属性
    public int no;
    public String name;
    public String nickname;
    public HeroNode next;
    // 3. 添加构造方法，初始化属性
    public HeroNode(int no, String name, String nickname) {
        this.no = no;
        this.name = name;
        this.nickname = nickname;
    }
    // 4. toString方法，显示节点信息

    @Override
    public String toString() {
        return "HeroNode{" +
                "no=" + no +
                ", name='" + name + ''' +
                ", nickname='" + nickname + ''' +
                '}';
    }
}

class SingleLinkedList {
    // 5. 创建一个单链表的类，用来添加对单链表的操作
    // 6. 添加节点
    private HeroNode head = new HeroNode(0, "", "");
    public HeroNode getHead() {
        return this.head;
    }
    public void add(HeroNode heroNode) {
        HeroNode temp = head;
        while (temp.next != null) {
            temp = temp.next;
        }
        temp.next = heroNode;
    }
    // 7. 有序添加节点
    public void addByOrder(HeroNode heroNode) {
        HeroNode temp = head;
        while (temp.next != null) {
            if (temp.next.no < heroNode.no) {
                temp = temp.next;
            }else if (temp.next.no > heroNode.no) {
                heroNode.next = temp.next;
                break;
            }else {
                System.out.printf("编号为 %d 的节点已经存在，添加失败~~~n", heroNode.no);
                return;
            }
        }
        temp.next = heroNode;
    }
    // 8. 显示单链表的节点信息
    public void list() {
        if (head.next == null) {
            System.out.println("该单链表为空，没有可以显示的节点信息~~~");
            return;
        }
        HeroNode temp = head;
        while (temp.next != null) {
            System.out.println(temp.next.toString());
            temp = temp.next;
        }
    }
}

尚硅谷04 数据结构与算法

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