鏈表操作-leetcode23-刪除倒數第幾個節點

給你一個鏈表，刪除鏈表的倒數第 n 個結點，並且返回鏈表的頭結點。

示例 1：

示例 2：

輸入：head = [1], n = 1
輸出：[]

示例 3：

輸入：head = [1,2], n = 1
輸出：[1]

提示：

• 鏈表中結點的數目爲 sz
• 1 <= sz <= 30
• 0 <= Node.val <= 100
• 1 <= n <= sz

思路：首先遍歷所有節點計算總數，然後計算出要跳過的那個節點的前一個節點，然後利用探測指針跳過，由於要跳過的可能是頭結點，所以需要假如一個探測節點，所以計算位置的時候 int position = len - n+1 -1+1;//要刪除的節點的前一個節點,因爲添加了一個虛節點，所以在+1 
需要這樣結算

//leetcode submit region begin(Prohibit modification and deletion)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 * int val;
 * ListNode next;
 * ListNode() {}
 * ListNode(int val) { this.val = val; }
 * ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if (head == null) {
            return null;
        }

        int len = 0;
        ListNode dummy = head;
        while (dummy != null) {
            len++;
            dummy = dummy.next;
        }

        int position = len - n+1 -1+1;//要刪除的節點的前一個節點,因爲添加了一個虛節點，所以在+1
        System.out.println("position: " + position);

        int currPosition = 0;
        ListNode dummy3 = new ListNode(-1);
        dummy3.next = head;
        ListNode dummy2 = dummy3;

        while (dummy2 != null) {
            currPosition++;
            System.out.println("currPosition: " + currPosition);

            if (currPosition == position) {
                System.out.println("dummy2.next: "+dummy2.next.val);
                //System.out.println("dummy2.next.next: "+dummy2.next.next.val);

                dummy2.next = dummy2.next.next;


                return dummy3.next;
            } else {
                dummy2 = dummy2.next;
            }

        }
        return null;

    }
}
//leetcode submit region end(Prohibit modification and deletion)