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.
基础题，考验对链表的理解。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode ln = null;
int sum = 0;
ListNode temp = head;
while (temp != null){
sum++;
temp = temp.next;
}
if(sum == 0 || n>sum || n<=0){
return head;
}
ln = head;
int count = sum - n;
if(count >0){
while(--count > 0){
ln = ln.next;
if(ln == null){
return head;
}
}
ln.next = ln.next.next;
}else{
head = head.next;
}
return head;
}
}