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Leetcode: Remove Nth Node From End of List

时间:2014-11-17 22:45:53      阅读:164      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.

分析:two pointers.

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        if(head == NULL) return head;
        
        ListNode *dummy = new ListNode(-1);
        dummy->next = head;
        
     //move fast pointer to (n+1)th node from the head ListNode
*fast = head; for(int i = 0; i < n; i++) fast = fast->next; //move fast to the end of the list ListNode *prev = dummy, *p = head; while(fast){ fast = fast->next; prev = p; p = p->next; } prev->next = p->next; return dummy->next; } };

 

Leetcode: Remove Nth Node From End of List

标签:style   blog   io   color   sp   for   div   on   log   

原文地址:http://www.cnblogs.com/Kai-Xing/p/4104515.html

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