19. Remove Nth Node From End of List

[fockspaces]

https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/

[fockspaces]

1. 用快慢指針，快指針提前慢指針走 n 個節點，之後兩者同速前進，當 fast 到最後一個節點時，slow 即為從最後數來第 N 個 node
2. 這邊要考慮 edge case：刪掉的 node 為 head，當 fast 在提前走時碰到 NULL，代表要刪的是頭，這邊直接 return head->next

   /**
   * Definition for singly-linked list.
   * struct ListNode {
   *     int val;
   *     ListNode *next;
   *     ListNode() : val(0), next(nullptr) {}
   *     ListNode(int x) : val(x), next(nullptr) {}
   *     ListNode(int x, ListNode *next) : val(x), next(next) {}
   * };
   */
   class Solution {
   public:
   ListNode* removeNthFromEnd(ListNode* head, int n) {
       ListNode *slow = head, *fast = head;
       while(n--) fast = fast->next;
       if(!fast) return head->next;
       while(fast && fast->next) 
           {fast = fast->next; slow = slow->next;}
       slow->next = slow->next ? slow->next->next : NULL;
       return head;
   }
   };