Leetcode: 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.

注意边界条件即可。

/**
* 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 *prev = head;
ListNode *tail = head;
while (n-- > 0) {
tail = tail->next;
}
while (tail != NULL && tail->next != NULL) {
tail = tail->next;
prev = prev->next;
}
if (tail == NULL) {
head = head->next;
}
else {
prev->next = prev->next->next;
}

return head;
}
};