From 2ecb3669d05cff13b76a162b0185055b8cfecd79 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <14021051@buaa.edu.cn>
Date: Tue, 18 Dec 2018 16:07:26 +0800
Subject: [PATCH] Create 19. Remove Nth Node From End of List.md

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+# [19. Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/)
+# 思路
+题意就是去掉链表中倒数第n个节点。   
+涉及到链表中倒数节点的思路都是使用两个指针p1、p2，两个指针初始都是head，p2先走n步，然后p1、p2再同时走直到p2到达链尾，
+此时p1就位于链表倒数第n个节点的前一个节点。     
+需要注意当要删除的节点就是head时需要特殊处理。    
+时间复杂度O(N)，空间复杂度O(1)
+
+# C++
+``` C++
+/**
+ * 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(n <= 0) return head;
+        ListNode *p1 = head, *p2 = head;
+        while(n--) p2 = p2 -> next;  // p2先走n步
+        if(p2 == NULL){ // 此时删掉的应该是head
+            head = head -> next;
+            delete p1;
+            return head;
+        }
+        while(p2 -> next){
+            p1 = p1 -> next;
+            p2 = p2 -> next;
+        } // 此时p1位于倒数第n个节点的前一个节点
+        p2 = p1 -> next;
+        p1 -> next = p2 -> next;
+        delete p2;
+        return head;
+    }
+};
+```