README.md

Remove Nth Node From End of List

Description

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

Tags: Linked List, Two Pointers

思路

题意是让你删除链表中的倒数第 n 个数，我的解法是利用双指针，这两个指针相差 n 个元素，当后面的指针扫到链表末尾的时候，自然它前面的那个指针所指向的下一个元素就是要删除的元素，即 pre.next = pre.next.next;，但是如果一开始后面的指针指向的为空，此时代表的意思就是要删除第一个元素，即 head = head.next;。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode pre = head;
        ListNode afterPreN = head;
        while (n-- != 0) {
            afterPreN = afterPreN.next;
        }
        if (afterPreN != null) {
            while (afterPreN.next != null) {
                pre = pre.next;
                afterPreN = afterPreN.next;
            }
            pre.next = pre.next.next;
        } else {
            head = head.next;
        }
        return head;
    }
}

结语

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