leetcode Remove Nth Node From End of List 2.7 难度系数2

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本文介绍了一种高效算法，用于从链表中删除倒数第N个节点，并返回修改后的链表头。该算法仅遍历一次链表，通过两个指针之间的距离来确定待删除节点的位置。

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Question:

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.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode base = new ListNode(0);
        base.next = head;
   
        if(n<1) return head;
        
        int i=n;
        ListNode p = base;
        ListNode q = p;
        while(i-->=0 && q!=null) {
            q=q.next;
        }
        while(q!=null) {
            q=q.next;
            p=p.next;
        }
        p.next = p.next.next;
        
        return base.next;
    
    }
}