Leet Code 19 Remove Nth Node From End of List

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.

【算法思路】

用两个指针，相差距离为n，先移动一个指针n次，然后再同时移动，当后一个指针为尾指针时，前一个指针指向了要删除元素的前一个节点。

【复杂度】

时间：O（n）

public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode deleteNode = head;
ListNode tailNode = head;

if((head.next == null) && (n >= 1))
{
return null;
}

for(int i = 0 ; i < n; i ++)
{
if(tailNode.next != null)
tailNode = tailNode.next;
else
{
if(i == n - 1)
head = head.next;
return head;
}
}

while(tailNode.next != null)
{
deleteNode = deleteNode.next;
tailNode = tailNode.next;
}

tailNode = deleteNode.next;
deleteNode.next = tailNode.next;
tailNode.next = null;
tailNode = null;

return head;
}