leetcode链表--18、remove-nth-node-from-end-of-list（从链表中删除倒数第k个结点）

题目描述

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

For 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.

Try to do this in one pass.

解题思路：参照找倒数第k个结点那个题

1）两个指针p1 p2  p2先走k-1，然后两个指针一起走，p2走到最后，p1即为倒数第k个结点pre为倒数第k+1个结点

2）从链表中删除第k个节点

本题应注意边界条件的判断：

1）链表为空

2）要删除的正好是第一个结点的情况

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *removeNthFromEnd(ListNode *head, int n) {
12         if(head == NULL)
13             return NULL;
14         ListNode *pPre = NULL;
15         ListNode *p1 = head;
16         ListNode *p2 = head;
17         for(int i=0;i<n-1;i++)
18         {
19             p2 = p2->next;
20         }
21         while(p2->next != NULL)
22         {
23             pPre = p1;
24             p1 = p1->next;
25             p2 = p2->next;
26         }
27         if (pPre == NULL)//正好要删除的就是第一个结点
28         {
29             head = p1->next;
30             delete p1;
31         }
32         else
33         {
34            pPre->next = p1->next;
35            delete p1;
36         }
37
38         return head;
39     }
40 };