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[LeetCode] Remove Nth Node From End of List

时间:2014-10-29 18:31:47      阅读:191      评论:0      收藏:0      [点我收藏+]

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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.

 

思路一:快慢指针。时间复杂度O(n),空间复杂度O(1)

 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 || n <= 0) return head;
13     ListNode *pfast = head;
14     for (int i = 1; i < n && pfast->next != NULL; ++i) {
15         pfast = pfast->next;
16     }
17     ListNode **pslow = &head;
18     while (pfast->next != NULL) {
19         pfast = pfast->next;
20         pslow = &((*pslow)->next);
21     }
22     
23     ListNode *q = *pslow;
24     *pslow = q->next;
25     delete q;
26     
27     return head;
28 }
29 };

 

[LeetCode] Remove Nth Node From End of List

标签:style   blog   io   color   for   sp   div   on   log   

原文地址:http://www.cnblogs.com/vincently/p/4060009.html

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