Leetcode Remove Nth Node From End of List

题意：删除链表中倒数第i个元素。

思路：用堆栈完成计数工作。

/**
* 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 head;
stack<ListNode*> mys;
ListNode* myhead = new ListNode(0);
ListNode* next = head;
myhead->next = next;
mys.push(myhead);
while(next) {
mys.push(next);
next = next->next;
}

ListNode* after = NULL;
next = mys.top();
mys.pop();
while(n --) {
after = next->next;
next = mys.top();
mys.pop();
}
next->next = after;

return myhead->next;
}
};

另一种只遍历一边的做法是，在链表中标记第i个元素到结尾的长度。这样做目的是记录从头到达i-1个元素需要的循环次数。
/**
* 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* myhead = new ListNode(0);
myhead->next = head;
ListNode* mark = head;
ListNode* next = myhead;
while(n --) {
mark = mark->next;
}

while(mark) {//cout << "here" <<endl;
mark = mark->next;
next = next->next;
}
next->next = next->next->next;

return myhead->next;
}
};