[LeetCode] 19. Remove Nth Node From End of List ☆

Given a linked list, remove the n^th 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.

解法：

　　题目要求移除倒数第n个节点，并且说明了n一定是有效的，限定一次遍历解决问题。首先定义两个指针 first 和 last，first 指向第一个节点，last 指向第 n+1 个节点；两个节点同时向后撸，直到 last.next 为 null 的时候，说明当前 first.next 即为待删除节点，直接 first.next.next = first.next 即可删除，然后返回 head。注意，如果一开始的时候 last 就为 null，说明 head 节点就是带删除的节点，直接返回 head.next 即可。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode first = head;
        ListNode last = head;
        while (n-- > 0) {
            last = last.next;
        }
        
        if (last == null) {
            return head.next;
        }
        
        while (last.next != null) {
            first = first.next;
            last = last.next;
        }
        
        first.next = first.next.next;
        return head;
    }
}