Leetcode42: 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.

定义一个指针q，q先走n步，在定义一个指针p，p和q同时走，直到q走到链表的结尾，这时p所指的元素就是要删除的元素，在定义一个指针指向p之前的元素，其next指向p的next即可。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode *p = head;
        ListNode *q = head;
        ListNode *pre = p;
        for(int i = 0; i < n; i++)
        {
            q = q->next;
        }
        while(q)
        {
            pre = p;
            p = p->next;
            q = q->next;
        }
        if(p == head)
        {
            head = head->next;
        }
        else
            pre->next = p->next;
        return head;
    }
};

Leetcode42: Remove Nth Node From End of List

原文：http://blog.csdn.net/u013089961/article/details/46693001