leetcode19. 删除链表的倒数第 N 个结点

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        /*
            通过一个指针p先走n-1步之后，另一个指针q从头开始走，那么当p移动到链表尾部时，q就会位于链表的倒数第n个节点。
        */
        ListNode p = head;
        for(int i = 0;i < n-1;i ++)
            p = p.next;
        ListNode q = head,pre = null;
        while(p != null && p.next != null){
            p = p.next;
            pre = q;
            q = q.next;
        }
        //pre为null,就说明当前p指针已经在链表尾了，要删除的是链表头部节点
        if(pre == null)
            head = head.next;
        else
            pre.next = q.next;
        return head;
    }
}