LeetCode(19)-Remove Nth Node From End of List

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19. Remove Nth Node From End of List

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.

嗯？这个题的意思就是给一个链表，然后删掉倒数的第某个位置，然后返回新链表，还是比较容易的
但是需要注意某些细节的地方。不想写了，睡觉。

/**
 * 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 h=head;
        int j=1;
        while (null!=(h=h.next))j++;
        if(j-n==0)return head.next;
        if(j==1)return null;
        h=head;
        for (int i =1;i<(j-n); i++) {
            if(null!=h.next)h=h.next;
            else return head;
        }
        if(h.next!=null)h.next=h.next.next;
        else h.next=null;
        return head;
    }
}