LeetCode 019 Remove Nth Node From End of List

题目描述：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.

分析：

题目中说n是合法的，就不用对n进行检查了。用标尺的思想，两个指针相距为n-1，后一个到表尾，则前一个到n了。

① 指针p、q指向链表头部；

② 移动q，使p和q差n-1；

③ 同时移动p和q，使q到表尾；

④ 删除p。

(p为second，q为first)

代码如下：

/**
 * 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 || head->next == NULL) return NULL;

        ListNode * first = head;
        ListNode * second = head;

        for(int i = 0;i < n;i++)
            first = first->next;

        if(first == NULL)
            return head->next;

        while(first->next != NULL){
            first = first->next;
            second = second->next;
        }

        second->next = second->next->next;
        return head;
    }
};

Java:

    public ListNode removeNthFromEnd(ListNode head, int n) {

        if (head == null || head.next == null) {
            return null;
        }

        ListNode first = head;
        ListNode second = head;

        for (int i = 0; i < n; i++) {
            first = first.next;
        }

        if (first == null) {
            return head.next;
        }

        while (first.next != null) {
            first = first.next;
            second = second.next;
        }

        second.next = second.next.next;
        return head;
    }