Leetcode 19. 删除链表的倒数第N个节点

19. 删除链表的倒数第N个节点

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解题思路

利用双指针，将快指针提前前进 n + 1 个节点，然后同时移动快慢指针。当快指针指向 null 时，慢指针恰好指向需要删除的节点的前一节点。

注意需要设置哨兵节点，否则对于只有一个节点的链表，该算法无法处理。

 

Java代码

/**
 * 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) {
        ListNode guard = new ListNode(0);
        guard.next = head;
        ListNode fast = guard;
        ListNode slow = guard;
        for (int i = 0; i <= n; i++) fast = fast.next;
        while (fast != null) {
            fast = fast.next;
            slow = slow.next;
        }
        slow.next = slow.next.next;
        return slow == guard ? slow.next : head;
    }
}