Remove Nth Node From End of List

描述
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
设两个指针p; q，让q 先走n 步，然后p 和q 一起走，直到q 走到尾节点，删除p->next 即可。

class Solution {public:ListNode *removeNthFromEnd(ListNode *head, int n) {ListNode dummy(-1);dummy.next=head;ListNode *p = &dummy, *q = &dummy;for (int i = 0; i < n; i++) // q 先走n 步q = q->next;while(q->next) { // 一起走p = p->next;q = q->next;}ListNode *tmp = p->next;p->next = p->next->next;delete tmp;return dummy.next;}};