題目

分別實現兩個函式，分別可以刪除單鏈表和雙鏈表中倒數第K個節點。

思路

兩次遍歷連結串列，第一遍每移動一步，就讓K值減1；第二遍從頭開始遍歷連結串列，每移動一步K值加1，加到0就停止遍歷，此時移動到的節點就是要刪除節點的前一個節點。

程式碼實現

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 {
    /**
     * 單鏈表
     *
     * @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;
    }

    /**
     * 雙鏈表
     *
     * @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;
    }
}