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.
解题思路:
难度不大,注意下边界条件即可:
JAVA实现:
static public ListNode removeNthFromEnd(ListNode head, int n) {
if(n<=0)
return head;
ListNode ln=head;
int i=1;
while(ln.next!=null){
ln=ln.next;
i++;
}
if(i==n)
return head.next;
ln=head;
for(;i>n+1;i--)
ln=ln.next;
ln.next=ln.next.next;
return head;
}
Java for LeetCode 019 Remove Nth Node From End of List
原文:http://www.cnblogs.com/tonyluis/p/4473394.html