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.
思路：双指针，相差n
/**
* 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* pFirst = head, *pSecond = head;
for(int i = ; i < n; i++){
pFirst = pFirst->next;
}
if(pFirst == NULL){
return head->next;
}
while(pFirst->next){
pFirst = pFirst->next;
pSecond = pSecond->next;
}
pSecond->next = pSecond->next->next;
return head;
}
};