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

For example,

  1. Given linked list: 1->2->3->4->5, and n = 2.
  2.  
  3. After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Try to do this in one pass.

原题链接:https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/

题目:给定一个链表,从尾部删除第n个节点并返回新的链表。

思路:使用两个指针,fast 和 slow,他们的距离是n,于是fast到尾的时候,n所在的节点就是须要删除的节点。

  1. public ListNode removeNthFromEnd(ListNode head, int n) {
  2. ListNode slow = head, fast = head;
  3. if (head.next == null)
  4. return null;
  5. for (int i = 1; i <= n; i++)
  6. slow = slow.next;
  7. if (slow == null) {
  8. head = head.next;
  9. return head;
  10. }
  11. while (slow.next != null) {
  12. slow = slow.next;
  13. fast = fast.next;
  14. }
  15. fast.next = fast.next.next;
  16. return head;
  17. }
  18. // Definition for singly-linked list.
  19. public class ListNode {
  20. int val;
  21. ListNode next;
  22. ListNode(int x) { val = x; }
  23. }

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