快慢指针-----Remove Nth Node From End of List-优快云博客

本文介绍了一种高效的方法来移除链表中倒数第N个节点。通过使用两个指针（快指针和慢指针），在遍历一次链表的同时定位到目标节点并进行删除操作。

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Given a linked list, remove the nth node from the end of list and return its head.

Have you met this question in a real interview? Yes
Example
Given linked list: 1->2->3->4->5->null, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5->null.
Note
The minimum number of nodes in list is n.

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The first node of linked list.
     * @param n: An integer.
     * @return: The head of linked list.
     */
    ListNode removeNthFromEnd(ListNode head, int n) {
        // write your code here
        if(head == null || head.next == null)
            return null;

        ListNode faster = head;
        ListNode slower = head;

        for(int i = 0; i<n; i++)
            faster = faster.next;

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

        while(faster.next != null){
            slower = slower.next;
            faster = faster.next;
        }

        slower.next = slower.next.next;
        return head;
    }
}