[Leetcode]19. 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.

这题可以设置两个指针，开始都指向第一个节点，第一个指针先后移，移动到第n个节点时，两个指针一起向后移动，当第一个指针到达最后一个节点时，第二个指针所指的位置就是倒数第n个节点。

/** * 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) {        if (head == NULL)            return NULL;        ListNode *pre = NULL;        ListNode *p = head;        ListNode *q = head;        for(int i = 0; i < n - 1; i++)            q = q->next;        while(q->next)        {            pre = p;            p = p->next;            q = q->next;        }        if (pre == NULL)        {            head = p->next;            delete p;        }        else        {            pre->next = p->next;            delete p;        }        return head;    }};