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.
双指针，第一个指针比第二个指针快n个节点。注意边界情况。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode tail = head;
while (n > 0) {
// No need to delete
if (tail == null) {
return head;
}
tail = tail.next;
n--;
}
// The deleted one is head
if (tail == null) {
return head.next;
}
ListNode tmp = head;
while (tail.next != null) {
tmp = tmp.next;
tail = tail.next;
}
tmp.next = tmp.next.next;
return head;
}
}
19. Remove Nth Node From End of List
原文：http://www.cnblogs.com/shini/p/4499285.html