leetcode-Remove Nth Node From End of List

Question:

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.

Solution:

/**
* 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) {
if (head == NULL){
return NULL;
}
ListNode* p1,*p2;
p1 = p2 = head;
for(int i = 0 ; i < n;i++){
p2 = p2->next;//题目中明确n为有效值，所以可以这么做；否则，需要做判断
//if(p2 == NULL){
//   return NULL;
//}
}
if(p2 != NULL){//确定删除的数不为第一个
p2 = p2->next;
}
else{//删除第一个数，直接返回
return head->next;
}
while(p2 != NULL){
p1 = p1->next;
p2 = p2->next;
}
p1->next = p1->next->next;
return head;
}
};

总结：

复杂度为O(n)