LCR.P21[删除链表的倒数第N个结点]

方法一：遍历

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
/**
 * 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) {
        ListNode dummy = new ListNode(0, head), p = head;
        int length = 0;
        while (p != null) {
            p = p.next;
            ++length;
        }
        p = dummy;
        for (int i = 0; i < length - n; ++i) {
            p = p.next;
        }
        p.next = p.next.next;
        return dummy.next;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(1)$

方法二：快慢指针

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
/**
 * 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) {
        ListNode dummy = new ListNode(0, head), slow = dummy, fast = head;
        while (n-- > 0) {
            fast = fast.next;
        }
        while (fast != null) {
            slow = slow.next;
            fast = fast.next;
        }
        slow.next = slow.next.next;
        return dummy.next;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(1)$