数据结构入门到入土——链表（1）

虽然有这么多的链表的结构，但是我们重点掌握两种:

三，链表的实现

和上期顺序表一样，我们采用接口的方式实现：

接口部分：

// 1、无头单向非循环链表实现
public interface IList {
    //头插法
    public void addFirst(int data);
    //尾插法
    public void addLast(int data);
    //任意位置插入,第一个数据节点为0号下标
    public void addIndex(int index,int data);
    //查找是否包含关键字key是否在单链表当中
    public boolean contains(int key);
    //删除第一次出现关键字为key的节点
    public void remove(int key);
    //删除所有值为key的节点
    public void removeAllKey(int key);
    //得到单链表的长度
    public int size();
    //清空链表
    public void clear();
    //打印链表
    public void display();
}

接口方法实现：

public class AchieveList implements IList{
    static class ListNode {
        public int val;
        public ListNode next;
        public ListNode(int val) {
            this.val = val;
        }
    }

    public ListNode head;

    //创建一个链表
    public void crateList() {
        ListNode node1 = new ListNode(12);
        ListNode node2 = new ListNode(34);
        ListNode node3 = new ListNode(56);
        ListNode node4 = new ListNode(78);
        ListNode node5 = new ListNode(99);
        node1.next = node2;
        node2.next = node3;
        node3.next = node4;
        node4.next = node5;
        this.head = node1;
    }

    //头插法
    @Override
    public void addFirst(int data) {
        ListNode node =new ListNode(data);
        node.next = this.head;
        this.head = node;
    }

    //尾插法
    @Override
    public void addLast(int data) {
        ListNode cur = this.head;
        while (cur.next != null) {
            cur = cur.next;
        }
        ListNode node = new ListNode(data);
        cur.next = node;
    }

    //任意位置插入,第一个数据节点为0号下标
    @Override
    public void addIndex(int index, int data) {
        ListNode node = new ListNode(data);
        ListNode cur = this.head;
        ListNode prev = this.head;
        if (index == 0) {
            node.next = cur;
            this.head = node;
        } else {
            int count = 0;
            while (count != index) {
                cur = cur.next;
                count++;
            }
            count = 0;
            while ((count+1) != index) {
                prev = prev.next;
                count++;
            }
            node.next = cur;
            prev.next = node;
        }
    }

    //查找是否包含关键字key是否在单链表当中
    @Override
    public boolean contains(int key) {
        ListNode cur = this.head;
        while (cur != null) {
            if (cur.val == key) {
                return true;
            }
            cur = cur.next;
        }
        return false;
    }

    //删除第一次出现关键字为key的节点
    @Override
    public void remove(int key) {
        if (head == null) {
            return;
        }
        ListNode cur = this.head;
        ListNode prev = this.head;
        int count = 1;
        while (cur.val != key) {
            if (count == size()) {
                return;
            }
            cur = cur.next;
            count++;
        }
        if (count == 1) {
            head = null;
            head = prev.next;
        } else {
            while (prev.next.val != key) {
                prev = prev.next;
            }
            prev.next = cur.next;
        }
    }

    //删除所有值为key的节点
    @Override
    public void removeAllKey(int key) {
        ListNode cur = this.head;
        int count = 1;
        while (cur.next != null) {
            if (cur.val == key) {
                count++;
            }
            cur = cur.next;
        }
        for (int i = 0; i < count; i++) {
            remove(key);
        }
    }

    //得到单链表的长度
    @Override
    public int size() {
        int count = 1;
        ListNode cur = this.head;
        if (cur == null) {
            return 0;
        }
        while (cur.next != null) {
            cur = cur.next;
            count++;
        }
        return count;
    }

    @Override
    public void clear() {
        ListNode cur = this.head;
        while (cur.next != null) {
            head = null;
            head = cur.next;
            cur = cur.next;
        }
        head = null;
    }

    //打印链表
    @Override
    public void display() {
        ListNode cur = this.head;
        if (head == null) {
            System.out.println("[" + "]");
        } else {
            System.out.print("[");
            while (cur != null) {
                if (cur.next == null) {
                    System.out.println(cur.val + "]");
                } else {
                    System.out.print(cur.val + " ");
                }
                cur = cur.next;
            }
        }
    }

四，与链表有关的题目练习（1）

1.删除链表中等于给定值 val 的所有节点

class Solution {
        public ListNode removeElements(ListNode head,int val) {
            if (head == null) {
                return null;
            }
            ListNode cur = head.next;
            ListNode prev = head;
            while (cur != null) {
                if (cur.val == val) {
                    prev.next = cur.next;
                    cur = cur.next;
                } else {
                    cur = cur.next;
                    prev = prev.next;
                }
            }
            if (head.val == val) {
                head = head.next;
            }
            return head;
        }
    }

2.反转一个单链表

思路：

现有如下链表：

定义cur = head.next

将head.next指向null

定义变量curNext = cur.next

将cur的下一个节点设为head

令cur = curNext；curNext = cur.next;

后以此思路循环

class Solution {
        public ListNode reverseList(ListNode head) {
            if (head == null) {
                return null;
            }
            if (head.next == null) {
                return head;
            }
            ListNode cur = head.next;
            head.next = null;
            while (cur != null) {
                ListNode curNext = cur.next;
                cur.next = head;
                head = cur;
                cur = curNext;
            }
            return head;
        }
    }

3.给定一个带有头结点 head 的非空单链表，返回链表的中间结点。如果有两个中间结点，则返回第二个中间结点

思路：

定义如图两个变量fast和slow，且都等于head

令fast每次走两步，slow每次走一步。当fast==null或者fast.next === null时

此时slow的位置必定是中间节点

初：

移动一次：

移动两次：

末：

class Solution {
        public ListNode middleNode(ListNode head) {
            ListNode fast = head;
            ListNode slow = head;
            while (fast != null && fast.next != null) {
                fast = fast.next.next;
                slow = slow.next;
            }
            return slow;
        }
    }
}

4.输入一个链表，输出该链表中倒数第k个结点

假设有以下链表：

要求倒数第k个节点，我们可以先定义fast和slow两个位于头节点的变量

假如K为3，我们就先令fast先走K-1步

然后令fast和slow一起走同样的步数，直到fast.next为null

此时slow对应的位置就是所求的倒数第K个节点

public class Solution {
        public ListNode FindKthToTail(ListNode head,int k) {
            if (head == null) {
                return head;
            }
            if (k <= 0 && head != null ) {
                return null;
            }
            ListNode fast = head;
            ListNode slow = head;
            for (int i = 0; i < k-1; i++) {
                fast = fast.next;
                //处理K太大的问题
                if (fast == null) {
                    return null;
                }
            }
            while (fast.next != null) {
                fast = fast.next;
                slow = slow.next;
            }
            return slow;
        }
    }

5.将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的

思路：

先有以下两个有序链表需要被合并为一个有序链表

定义一个newH节点

将head1.val与head2.val进行比较，较小令它等于tmpH且令它等于它的下一个节点，tmpH为newH的下一个节点

如上图head1.val<head2.val,故发生以下变化

接着对head1.val与head2.val进行比较

head2.val更小

继续比较

head1.val更小

继续比较

head2.val更小

……

当其中有一个数组比较完了的时候，此时只需令tmpH.next为另一个head即可

最后返回newH.next即可

class Solution {
        public ListNode mergeTwoLists(ListNode head1, ListNode head2) {
            ListNode newH = new ListNode(-1);
            ListNode tmpH = newH;
            while (head1 != null && head2 != null) {
                if (head1.val < head2.val) {
                    tmpH.next = head1;
                    tmpH = tmpH.next;
                    head1 = head1.next;
                } else {
                    tmpH.next = head2;
                    tmpH = tmpH.next;
                    head2 = head2.next;
                }
            }
            if (head1 != null) {
                tmpH.next = head1;
            }
            if (head2 != null) {
                tmpH.next = head2;
            }
            return newH.next;
        }
    }

6.编写代码，以给定值x为基准将链表分割成两部分，所有小于x的结点排在大于或等于x的结点之前

使用cur遍历顺序表

当cur.val<给定的值x时就让它进入链表1

当cur.val>=给定的值x时就让它进入链表2

最后将链表2的尾部插在链表1的头部，返回链表2

public class Partition {
        public ListNode partition(ListNode pHead, int x) {
            ListNode cur = pHead;
            ListNode tmpH1 = new ListNode(-1);
            ListNode tmpH2 = new ListNode(-1);
            ListNode head1 = tmpH1;
            ListNode head2 = tmpH2;
            while (cur != null) {
                if (cur.val < x) {
                    head2.next = cur;
                    head2 = head2.next;
                } else {
                    head1.next = cur;
                    head1 = head1.next;
                }
                cur = cur.next;
            }
            tmpH2 = tmpH2.next;
            tmpH1 = tmpH1.next;
            head2.next = tmpH1;
            head1.next = null;
            if (tmpH2 == null) {
                return tmpH1;
            }
            if (tmpH1 == null) {
                return tmpH2;
            }
            return tmpH2;
        }
    }

7.链表的回文结构

通过以上快慢指针的方式找到链表的中间节点

翻转

当slow和fast相遇时停止

public class PalindromeList {
        public boolean chkPalindrome(ListNode A) {
            if (A == null || A.next == null) {
                return true;
            }
            ListNode fast = A;
            ListNode slow = A;
            //找中间位置
            while (fast != null && fast.next != null) {
                fast = fast.next.next;
                slow = slow.next;
            }
            ListNode cur = slow.next;
            //翻转
            while (cur != null) {
                ListNode curNext = cur.next;
                cur.next = slow;
                slow = cur;
                cur = curNext;
            }
            ListNode right = slow;
            ListNode left = A;
            //从前到后  从后到前
            while (left.next != right.next ) {
                if (left.val != right.val) {
                    return false;
                }
                if (left.next == right)
                {
                    return true;
                }
                left = left.next;
                right = right.next;
            }
            return true;
        }
    }