19. Remove Nth Node From End of List

Medium

Given a linked list, remove the n-th node from the end of list and return its head.

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.

虽然很久没有写数据结构的代码了，但是看完题目还是能顺手敲出来：）

结果一提交怎么都不对，原来啊，LeetCode的链表头结点也是存数据的，并不仅仅是头的作用

最后吐槽一下，真的很喜欢LeetCode这种只注重于算法过程的代码，不像我学校的OJ，有时候输入输出都会被卡住：）

/**
 * 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) {
        int len = 0;
        ListNode *a = head;
        while(a)
        {
            len++;
            a = a->next;
        }
        a = head;
        ListNode* b = new ListNode(0);
        b->next = a;
        for(int i = 1; i <= len - n; i++)
        {
            b = a;
            a = a->next;
        }
        b->next = a->next;
        if(len == n) return b->next;
        return head;
    }
};