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.
解题思路一：
先计算length，然后删除
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;
}
解题思路二：
一个指针先走n步，另一个指针跟上。
C++：
1 class Solution { 2 public: 3 ListNode* removeNthFromEnd(ListNode* head, int n) { 4 if (n <= 0) 5 return head; 6 ListNode* cur = head; 7 for (int i = 0; i < n-1; i++) { 8 if (cur->next != NULL) 9 cur = cur->next; 10 else return head; 11 } 12 if (cur->next == NULL) { 13 ListNode*temp = head; 14 head = head->next; 15 delete temp; 16 return head; 17 } 18 cur = cur->next; 19 ListNode* cur2 = head; 20 while (cur->next != NULL) { 21 cur2 = cur2->next; 22 cur = cur->next; 23 } 24 cur = cur2->next; 25 cur2->next = cur->next; 26 delete cur; 27 return head; 28 } 29 };