【LeetCode】019. Remove Nth Node From End of List

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.

 

题解:

  快慢指针。

 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)
13             return head;
14         ListNode dummy(-1);
15         dummy.next = head;
16         ListNode* fast = &dummy, *slow = &dummy;
17         
18         while (n-- >= 0 && fast) fast = fast->next;
19         
20         while(fast) {
21             fast = fast->next;
22             slow = slow->next;
23         }
24         
25         ListNode* tmp = slow->next;
26         slow->next = slow->next->next;
27         delete tmp;
28         
29         return dummy.next;
30     }
31 };

 

posted @ 2018-03-25 20:06  Vincent丶丶  阅读(141)  评论(0编辑  收藏  举报