[LeetCode] 19. Remove Nth Node From End of List

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

Example 1:

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]

Example 2:
Input: head = [1], n = 1
Output: []

Example 3:
Input: head = [1,2], n = 1
Output: [1]

Constraints:
The number of nodes in the list is sz.
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

Follow up: Could you do this in one pass?

移除链表倒数第N个节点。

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

思路

思路是在头结点之前（一般我们创建dummy节点的位置）创建两个节点分别叫做 slow 和 fast。先让 fast 前移 N + 1 个节点，这样使得 slow 和 fast 之间有 N + 1 个节点的距离。然后同时让 slow 和 fast 前移，直到 fast 遍历到链表尾部。此刻 slow 停下来的位置就是需要移除的节点之前的那个节点。为什么需要使 slow 和 fast 之间有 N + 1 个节点的距离是因为我们移除的是 slow.next 这个节点。可以参考这个动图帮助理解。

复杂度

时间O(n)
空间O(1)

代码

Java实现

/**
 * 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) {
        ListNode dummy = new ListNode(0);
        ListNode slow = dummy;
        ListNode fast = dummy;
        dummy.next = head;
        for (int i = 0; i <= n; i++) {
            fast = fast.next;
            // System.out.println(fast.val);
        }
        while (fast != null) {
            fast = fast.next;
            slow = slow.next;
        }
        slow.next = slow.next.next;
        return dummy.next;
    }
}

JavaScript实现

/**
 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
var removeNthFromEnd = function(head, n) {
    let dummy = new ListNode(0);
    let slow = dummy;
    let fast = dummy;
    dummy.next = head;
    for (let i = 0; i <= n; i++) {
        fast = fast.next;
    }
    while (fast !== null) {
        slow = slow.next;
        fast = fast.next;
    }
    slow.next = slow.next.next;
    return dummy.next;
};