LeetCode | #19 Remove Nth Node From End of List

题目：

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.

思路：

先算节点个数count，把顺数第count-n个数的next改为第count-n-2个数；
题目要求最好一遍遍历，那就靠数学的巧妙的，两指针，first先走n步，再first和second同时走，直到first的next是空，就把second的next改为next的next

public class RemoveNthFromEnd {

public class ListNode {
int val;
ListNode next;
ListNode(int x) {
val = x;
next = null;
}
}
/*//先算节点个数count，把顺数第count-n个数的next改为第count-n-2个数
public ListNode removeNthFromEnd(ListNode head, int n) {

int count = 1;
ListNode temp = head;
while(temp.next != null){
count++;
temp = temp.next;
}
int i = count-n;
if(i == 0){
return head.next;
}
temp = head;
while(i > 1){
temp = temp.next;
i--;
}
temp.next = temp.next.next;

return head;
}*/
//题目要求最好一遍遍历，那就靠数学的巧妙的，两指针，first先走n步，
//再first和second同时走，直到first的next是空，就把second的next改为next的next
public ListNode removeNthFromEnd(ListNode head, int n) {

ListNode first = head, second = head;
for(int i=n; i>0; i--){
first = head.next;
}
if(first == null){
return first;
}
while(first.next != null){
first = first.next;
second = second.next;
}
second.next = second.next.next;

return head;
}
}