LeetCode Remove Nth Node From End of List

class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n) {
if (head == NULL) return NULL;
ListNode* pre = NULL;
ListNode* cur = head;
ListNode* fast= cur;

for (int i=1; i<n; i++) {
fast = fast->next;
}

while (fast->next != NULL) {
pre = cur;
cur = cur->next;
fast= fast->next;
}

if (pre == NULL) {
return head->next;
} else {
pre->next = cur->next;
return head;
}
}
};

一次过呗

第二轮：

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) {}
* };
*/
// 10:16
class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n) {
ListNode fakeHead(0);
fakeHead.next = head;
ListNode* pre = NULL;
ListNode* slow = &fakeHead;
ListNode* fast = &fakeHead;

int k = n;
while (k--) {
fast = fast->next;
}

while (fast != NULL) {
fast = fast->next;
pre = slow;
slow = slow->next;
}
pre->next = slow->next;
return fakeHead.next;
}
};