Java版：LeetCode 19. Remove Nth Node From End of List

LeetCode 19. Remove Nth Node From End of List

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

难度：Medium

题目描述：

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

输入样例:

Example 1:

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.

该题已提供代码：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        
    }
}

解题思路（快慢指针思想）

1. 初始化一个DummyNode，让快慢指针都指向它，需要删除第n个节点，就将该节点的前驱直接指向该节点的后继即可。

2. 先将快指针移动到第n的位置上。

3. 再将双指针同时移动。

实现代码

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummyNode = new ListNode(0);
        dummyNode.next = head;
        ListNode slow = dummyNode;
        ListNode fast = dummyNode;
        for (int i = 0; i < n; i++) {
            fast = fast.next;
        }
        while (fast.next != null){
            slow = slow.next;
            fast = fast.next;
        }
        slow.next = slow.next.next;
        return dummyNode.next;
    }
}

LeetCode测试结果

结语

至此，Remove Nth Node From End of List 解决完毕！

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