19. Remove Nth Node From End of List(移除倒数第N的结点， 快慢指针)

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.

思路：

利用快指针先走n步，找到倒数第n个节点

然后删除。

 

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode() : val(0), next(nullptr) {}
 7  *     ListNode(int x) : val(x), next(nullptr) {}
 8  *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 9  * };
10  */
11 class Solution {
12 public:
13     ListNode* removeNthFromEnd(ListNode* head, int n) {
14         ListNode* slow = head;
15         ListNode* fast = head;
16         for(;n > 0;--n) {
17             fast = fast->next;
18         }
19         //如果此时快指针走到头了，// 说明倒数第 n 个节点就是第一个节点
20         if (fast == nullptr) return head->next;
21         while(fast != nullptr && fast->next!=nullptr) {
22             fast = fast->next;
23             slow = slow->next;
24         }
25         slow->next = slow->next->next;
26         return head;
27 
28     }
29 };