19. Remove Nth Node From End of List

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.

这题细节很多

1.首先删除倒数第n个节点，即我们要找到倒数第n个节点前面的那个节点，所以

while (n){
faster = faster->next;
n--;
}
while (faster->next){
faster = faster->next;
slower = slower->next;
}2.针对特殊情况，如1->NULL ,1要删除的就是头节点。
我们加一个不带内容的头结点，这样操作就可以统一，而且

ListNode* faster = &dummy;
ListNode* slower = &dummy;

/**
* 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 dummy(0);
dummy.next = head;
ListNode* faster = &dummy;
ListNode* slower = &dummy;
while (n){
faster = faster->next;
n--;
}
while (faster->next){
faster = faster->next;
slower = slower->next;
}
ListNode* d = slower->next;
ListNode* next2 = d->next;
slower->next = next2;
delete d;
return dummy.next;
}
};