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.
采用递归的方法实现one pass
public class Solution {
int cnt=0, target=0;
public ListNode removeNthFromEnd(ListNode head, int n) {
target = n;
return removeNthFromEnd(head);
}
private ListNode removeNthFromEnd(ListNode head){
if(head==null){
cnt++;
return null;
}
head.next = removeNthFromEnd(head.next);
return cnt++==target? head.next : head;
}
}
利用两个指针实现的方法
A one pass solution can be done using pointers. Move one pointer fast --> n+1 places forward, to maintain a gap of n between the two pointers and then move both at the same speed. 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);
ListNode slow = start, fast = start;
slow.next = head;
//Move fast in front so that the gap between slow and fast becomes n
for(int i=1; i<=n+1; i++) {
fast = fast.next;
}
//Move fast to the end, maintaining the gap
while(fast != null) {
slow = slow.next;
fast = fast.next;
}
//Skip the desired node
slow.next = slow.next.next;
return start.next;
}