LeetCode OJ - 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.

/**
* 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) return NULL;
if(!head->next) return NULL;
ListNode dummy(0);
dummy.next = head;

ListNode *last = &dummy;
ListNode *slow = dummy.next;
ListNode *fast = dummy.next;
int index = 1;
while(index < n) {
fast = fast->next;
index++;
}

while(fast->next) {
fast = fast->next;
last = slow;
slow = slow->next;
}

last->next = slow->next;
return dummy.next;
}
};

上面代码用fast->next让fast指针定位到了最后一个节点，让fast定位到最后的null指针也可以，但是前面fast要比slow多走两步。

/**
* 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) return NULL;
if(!head->next) return NULL;
ListNode dummy(0);
dummy.next = head;

ListNode *last = &dummy;
ListNode *slow = dummy.next;
ListNode *fast = dummy.next;
for(int i = 0; i < n; i++)
fast = fast->next;
while(fast) {
fast = fast->next;
last = slow;
slow = slow->next;
}

last->next = slow->next;
return dummy.next;
}
};