LeetCode 19. 删除链表的倒数第N个节点

package 链表;

/**
 * https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/
 * 19. 删除链表的倒数第N个节点
 * 思路：采用快慢指针，让快指针先走n步，慢指针从头开始，如果快指针的下一个节点为null，
 * 说明此时慢指针当前的节点就是要删除的节点，需要将慢指针指向下下个节点(跳过下一个节点)
 */
public class _19_Remove_Nth_Node_From_End_of_List {

    class ListNode {
        int val;
        ListNode next;

        ListNode(int x) {
            val = x;
            next = null;
        }
    }

    class Solution {
        public ListNode removeNthFromEnd(ListNode head, int n) {
            ListNode fast = head;
            while (n-- > 0) {
                fast = fast.next;
            }
            if (fast == null) {
                return head.next;
            }
            ListNode slow = head;
            while (fast.next != null) {
                fast = fast.next;
                slow = slow.next;
            }
            slow.next = slow.next.next;
            return head;
        }
    }
}

LeetCode 19. 删除链表的倒数第N个节点

原文：https://www.cnblogs.com/jianzha/p/12838029.html