算法问题(链表)--反转链表II

LeetCode第92号题–反转链表II

题目如下

反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。

说明:
1 ≤ m ≤ n ≤ 链表长度。

示例:

输入: 1->2->3->4->5->NULL, m = 2, n = 4
输出: 1->4->3->2->5->NULL

解答

public static ListNode reverseBetween(ListNode head, int m, int n) {
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;
        ListNode pre = dummyHead;
        //找到m前的那个结点
        for (int i = 1; i < m; i++)
            pre = pre.next;
        //head指向需要更换的第一个数
        head = pre.next;
        for (int i = m; i < n; i++) {
            //累计将需要更换的前1,2的结点更换
            //pre->head->next->next.next ===> pre->next->head->next.next
            ListNode next = head.next;
            head.next = next.next;
            next.next = pre.next;
            pre.next = next;
        }
        return dummyHead.next;
    }

因为链表算法不好测试，所以我们自己设计链表来测试，根据此题，全部代码如下：

class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {
        val = x;
    }

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

public class LinkedList {

    private ListNode dummyHead;
    private int size;

    public LinkedList() {
        dummyHead = new ListNode(0);
        size = 0;
    }

    // 获取链表中的元素个数
    public int getSize() {
        return size;
    }

    // 返回链表是否为空
    public boolean isEmpty() {
        return size == 0;
    }

    // 在链表的index(0-based)位置添加新的元素e
    public void add(int index, int val) {

        if (index < 0 || index > size)
            throw new IllegalArgumentException("Add failed. Illegal index.");

        ListNode prev = dummyHead;
        for (int i = 0; i < index; i++)
            prev = prev.next;

        prev.next = new ListNode(val, prev.next);
        size++;
    }

    // 在链表头添加新的元素val
    public void addFirst(int val) {
        add(0, val);
    }

    // 在链表末尾添加新的元素val
    public void addLast(int val) {
        add(size, val);
    }

    // 获得链表的第index(0-based)个位置的元素
    // 在链表中不是一个常用的操作，练习用：）
    public ListNode get(int index) {

        if (index < 0 || index >= size)
            throw new IllegalArgumentException("Get failed. Illegal index.");

        ListNode cur = dummyHead.next;
        for (int i = 0; i < index; i++)
            cur = cur.next;
        return cur;
    }

    // 获得链表的第一个元素
    public ListNode getFirst() {
        return get(0);
    }

    // 获得链表的最后一个元素
    public ListNode getLast() {
        return get(size - 1);
    }

    // 修改链表的第index(0-based)个位置的元素为val
    public void set(int index, int val) {
        if (index < 0 || index >= size)
            throw new IllegalArgumentException("Set failed. Illegal index.");

        ListNode cur = dummyHead.next;
        for (int i = 0; i < index; i++)
            cur = cur.next;
        cur.val = val;
    }

    // 查找链表中是否有元素val
    public boolean contains(int val) {
        ListNode cur = dummyHead.next;
        while (cur != null) {
            if (cur.val == val)
                return true;
            cur = cur.next;
        }
        return false;
    }

    // 从链表中删除index(0-based)位置的元素, 返回删除的元素
    public int remove(int index) {
        if (index < 0 || index >= size)
            throw new IllegalArgumentException("Remove failed. Index is illegal.");

        ListNode prev = dummyHead;
        for (int i = 0; i < index; i++)
            prev = prev.next;

        ListNode removeNode = prev.next;
        prev.next = removeNode.next;
        removeNode.next = null;
        size--;

        return removeNode.val;
    }

    // 从链表中删除第一个元素, 返回删除的元素
    public int removeFirst() {
        return remove(0);
    }

    // 从链表中删除最后一个元素, 返回删除的元素
    public int removeLast() {
        return remove(size - 1);
    }

    // 从链表中删除元素val
    public void removeElement(int val) {

        ListNode prev = dummyHead;
        while (prev.next != null) {
            if (prev.next.val == val)
                break;
            prev = prev.next;
        }

        if (prev.next != null) {
            ListNode delNode = prev.next;
            prev.next = delNode.next;
            delNode.next = null;
            size--;
        }
    }

    @Override
    public String toString() {
        StringBuilder res = new StringBuilder();
        ListNode cur = dummyHead.next;
        while (cur != null) {
            res.append(cur.val + "->");
            cur = cur.next;
        }
        res.append("NULL");

        return res.toString();
    }

    public static String print(ListNode node) {
        StringBuilder res = new StringBuilder();
        while (node != null) {
            res.append(node.val + "->");
            node = node.next;
        }
        res.append("NULL");
        return res.toString();
    }

    public static ListNode reverseBetween(ListNode head, int m, int n) {
        ListNode dummyHead = new ListNode(-1);
        dummyHead.next = head;
        ListNode pre = dummyHead;
        //找到m前的那个结点
        for (int i = 1; i < m; i++)
            pre = pre.next;
        //head指向需要更换的第一个数
        head = pre.next;
        for (int i = m; i < n; i++) {
            //累计将需要更换的前1,2的结点更换
            //pre->head->next->next.next ===> pre->next->head->next.next
            ListNode next = head.next;
            head.next = next.next;
            next.next = pre.next;
            pre.next = next;
        }
        return dummyHead.next;
    }

    public static void main(String[] args) {
        LinkedList linkedList = new LinkedList();
        for (int i = 0; i < 5; i++) {
            linkedList.addLast(i);
//            System.out.println(linkedList);
        }
        ListNode node = linkedList.get(0);
        System.out.println(print(node));

        ListNode node1 = reverseBetween(node,2,4);
        System.out.println(print(node1));
    }
}

此题是LeetCode第206号题–反转链表的升级版