[LeetCode OJ]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.

开始时候自己想是要从后往前找，想得到个长度，不过后来又一想，链表怎么size。。。

还是双指针靠谱，一前一后，后面的比前面的大n-1。

/** * 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 *temp = NULL;        ListNode *first = head;        ListNode *second = head;                for(int i = 1; i < n; i++)            first = first->next;                while(first->next) {            temp = second;            first = first->next;            second = second->next;        }                if(temp == NULL) {            head = second->next;            delete second;        }                else {            temp->next = second->next;            delete second;        }        return head;    }};