Remove Nth Node From End of List（19）
Given a linked list, remove the n-th node from the end of list and return its head.
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.
Follow up:
Could you do this in one pass?
关键技巧，快慢指针：
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public static ListNode removeNthFromEnd(ListNode head, int n) { //head 1-2-3-4-5
if (head == null || n <= 0) {
return head;
}
ListNode guide = new ListNode(-1);
guide.next = head;
ListNode first = guide;
ListNode second = guide;
for (int i = 0; i < n; i++) {
if (first != null) {
first = first.next; //1-2-3-4-5 //2-3-4-5 //3-4-5 //4-5
}
}
while (first != null && first.next != null){
first = first.next;
second = second.next;
}
second.next = second.next.next;
return guide.next;
}
}
