LeetCode 19. Remove Nth Node From End of List

LeetCode 19. Remove Nth Node From End of List
Solution1：我的答案
并不算是最优解法。

/**
 * 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 || n == 0)
            return NULL;
        struct ListNode* dummy = new ListNode(-1);
        dummy->next = head;
        struct ListNode* cur = dummy;
        int count = 0;
        while (cur->next) {count++; cur = cur->next;}
        count = count - n + 1;
        cur = dummy;
        while (count-- > 1) cur = cur->next;
        cur->next = cur->next->next;
        return dummy->next;
    }
};

Solution2：
参考网址：http://www.cnblogs.com/grandyang/p/4606920.html
其实这是一道经典的链表双指针算法题。
需要用两个指针来帮助我们解题，pre和cur指针。首先cur指针先向前走N步，如果此时cur指向空，说明N为链表的长度，则需要移除的为首元素，那么此时我们返回head->next即可，如果cur存在，我们再继续往下走，此时pre指针也跟着走，直到cur为最后一个元素时停止，此时pre指向要移除元素的前一个元素，我们再修改指针跳过需要移除的元素即可。代码如下：

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (!head->next) return NULL;
        ListNode *pre = head, *cur = head;
        for (int i = 0; i < n; ++i) cur = cur->next;
        if (!cur) return head->next;
        while (cur->next) {
            cur = cur->next;
            pre = pre->next;
        }
        pre->next = pre->next->next;
        return head;
    }
};