Given a linked list, remove the n-th node from the end of list and return its head.
Example:
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| 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.
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Note:
Given n will always be valid.
Follow up:
Could you do this in one pass?
给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点。
示例：
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| 给定一个链表: 1->2->3->4->5, 和 n = 2.
当删除了倒数第二个节点后，链表变为 1->2->3->5.
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说明：
给定的 n 保证是有效的。
进阶：
你能尝试使用一趟扫描实现吗？
想法
快慢指针，快指针指向第K个节点，慢指针指向开始节点，当快指针到达末尾时慢指针正好指向倒数第K的节点。由于要删除该节点，还得记录慢指针的上一个节点。Corner Case：传入参数为空、删除的是头节点、K大于链表长度。
解
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class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n)
{
if (head == NULL) {
return head;
}
ListNode *cur = head, *ans = head, *last = NULL;
int i = n - 1;
while (cur->next != NULL) {
if (i <= 0) {
last = ans;
ans = ans->next;
}
else {
i--;
}
cur = cur->next;
}
if (last == NULL) {
ListNode *temp = head;
head = head->next;
delete (temp);
}
else {
last->next = ans->next;
delete (ans);
}
return head;
}
};
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