Given the head of a linked list, remove the nth node from the end of the list and return its head.
Follow up: Could you do this in one pass?
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
Solution:
two pointer technique
- set slow pointer and fast pointer
- distance between two pointer is n
- until fast.next reach null
- slow.next is value want to remove
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| /**
* 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 start = new ListNode(0);
ListNode slow = start;
ListNode fast = start;
start.next = head;
while (n >= 0) {
fast = fast.next;
n--;
}
while (fast != null) {
slow = slow.next;
fast = fast.next;
}
slow.next = slow.next.next;
return start.next;
}
}
|
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