2.2在单链表和双链表中删除倒数第K个节点

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题目

分别实现两个函数，分别可以删除单链表和双链表中倒数第K个节点。

思路

两次遍历链表，第一遍每移动一步，就让K值减1；第二遍从头开始遍历链表，每移动一步K值加1，加到0就停止遍历，此时移动到的节点就是要删除节点的前一个节点。

代码实现

class Node {
    public int value;
    public Node next;

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

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

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

public class RemoveLastKthNode {
    /**
     * 单链表
     *
     * @param head
     * @param lastKth
     * @return
     */
    public Node removeLastKthNode(Node head, int lastKth) {
        if (head == null || lastKth < 1) {
            return head;
        }

        Node cur = head;

        while (cur != null) {
            lastKth--;
            cur = cur.next;
        }
        if (lastKth == 0) {
            head = head.next;
        }
        if (lastKth < 0) {
            cur = head;
            while (++lastKth != 0) {
                cur = cur.next;
            }
            cur.next = cur.next.next;
        }

        return head;
    }

    /**
     * 双链表
     *
     * @param head
     * @param lastKth
     * @return
     */
    public DoubleNode removeLastKthDoubleNode(DoubleNode head, int lastKth) {
        if (head == null || lastKth < 1) {
            return head;
        }

        DoubleNode cur = head;

        while (cur != null) {
            lastKth--;
            cur = cur.next;
        }
        if (lastKth == 0) {
            head = head.next;
            head.last = null;
        }
        if (lastKth < 0) {
            cur = head;
            while (++lastKth != 0) {
                cur = cur.next;
            }
            DoubleNode newNext = cur.next.next;
            cur.next = newNext;
            if (newNext != null) {;
                newNext.last = cur;
            }
        }

        return head;
    }
}