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.
这道题是个单链表的节点删除问题,值得注意的地方我标了红字,只扫描一次完成。这里需要引入三个指针trail、pos、pre_pos 。trail这个指针指示尾部,直到trail -> next为NULL的时候,就到了链表的结尾,关键之处在于,让trail先走n - 1个位置之后,pos再开始与trail同步前进,这样前后两个指针相差n - 1个元素,所以当trail到了结尾的时候,pos指示的位置就是倒数第n个位置。而pre_pos是用于标记pos之前一个元素,用来删除pos。
注意:最好在最后用delete把需要删的节点空间释放了,虽然这个和能不能AC没关系。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n) {
ListNode *trail = head, *pos = head, *pre_pos = NULL;
for (int i = 0; i < n - 1; i++){
trail = trail -> next;
}
if (trail -> next == NULL)
return head -> next;
while (trail -> next != NULL){
pre_pos = pos; //这里用这个隐含的表达了pre_pos延迟pos一个位置
trail = trail -> next;
pos = pos -> next;
}
pre_pos -> next = pos -> next;
delete pos;
return head;
}
};