Given a linked list, remove the nth node from the end of list and return its head.
For example,
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.
Try to do this in one pass.
[分析] 快慢指针思路。两指针初始均指向head,快指针尽量先走 n 步:
1)若快指针超过了最后一个节点,即为null:
a)若此时已走过n步,说明共有 n 个节点,需要删除第一个节点;
b)若此时尚未走完n步,说明链表长度小于n,无需删除任何节点
2)若快指针走完n步且没有超过最后一个节点,说明链表长度大于n,两指针开始同步前进,
快指针走到最后一个节点时,慢指针的下一个节点即为倒数第 n 个节点。
这题虽然为简单题,可是我并不能很快地做到bug free。
public class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { if (n <= 0) return head; ListNode slow = head, fast = head; int i = 0; while(i < n && fast != null) { fast = fast.next; i++; } if (fast == null) return i < n ? head : head.next; while (fast.next != null) { slow = slow.next; fast = fast.next; } slow.next = slow.next.next; return head; } }