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.
** 解題報(bào)告 **
使用快慢指針來(lái)做。
Using fast and slow pointers.
(1) Move one pointer fast --> n+1 places forward, to maintain a gap of n between the two pointers
(2) and then move both at the same speed.
(3) Finally, when the fast pointer reaches the end, the slow pointer will be n+1 places behind - just the right spot for it to be able to skip the next node.
Since the question gives that n is valid, not too many checks have to be put in place. Otherwise, this would be necessary.
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode start = new ListNode(0);
start.next = head;
ListNode fast = start;
ListNode slow = start;
// Move fast in front so that the gap between fast and slow becomes n;
for(int i = 1; i <= n + 1; i++) {
fast = fast.next;
}
// Moving fast to the end, maintaining the gap
while(fast != null) {
fast = fast.next;
slow = slow.next;
}
// Skip the desired node
slow.next = slow.next.next;
return start.next;
}
