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.

刪除倒數第n個節點，主要有兩種做法

法一：先遍歷得到鏈表的長度length，倒數第n個就是正數第length-n+1個（如1 2 3，倒數第三個節點就是3-3+1=1，即第一個節點）

/**
 * 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* newlist= new ListNode(0);
        newlist->next = head;  //帶頭節點的鏈表
        int length = 0;
        ListNode* first = head;
        while(first != NULL){   //得到鏈表長度
            length++;
            first = first->next;
        }
        if(n > length) return NULL;
        
 
int n1 = length + 1 - n;//倒數第n個是在正數第n1個節點
        first = newlist; 
        while(n1 - 1 > 0){         //找到目標節點的前一個位置，便於刪除目標節點
            n1--;
            first = first->next;
        }
        first->next = first->next->next;  //刪除目標節點
        return newlist->next;  //返回不帶頭節點的鏈表
    }
};

法二：利用兩個指針，快指針先走，找到第n個節點，再啟動慢指針。

此時快慢指針相距n，讓兩個指針一起走，快指針走到尾節點時，慢指針剛好到倒數第n個節點。

/**
 * 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)
            return nullptr;

        ListNode new_head(-1);
        new_head.next = head;

        ListNode *slow = &new_head, *fast = &new_head;

        for (int i = 0; i < n; i++)  //快指針先走到第n個節點
            fast = fast->next;

        while (fast->next)   //快慢指針一起走
        {
            fast = fast->next;
            slow = slow->next;
        }

        ListNode *to_do_deleted = slow->next;   //這兩行負責刪除操作
        slow->next = slow->next->next;

        delete to_do_deleted;

        return new_head.next;
    }
};

19. Remove Nth Node From End of List 刪除倒數第n各節點