快慢指针 03

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https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/

快慢指针解法

快指针先跑n个节点，如果跑到null了，说明此时慢指针就在倒数第n的位置，直接删除，即返回head的下一个元素即可。

没跑到null，则两指针一起跑，直到fast跑到null，然后通过slow删掉节点，返回head即可

/**
 * 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) {
        if (head == null || head.next == null) return null;
        ListNode fast = head, slow = head;

        for (int i = 0; i < n; i++) {
            fast = fast.next;
        }
        if (fast == null) {
            return head.next;
        }

        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        slow.next = slow.next.next;
        return head;
    }
}