feat: add 019

Author: Blankj

README.md

@@ -12,6 +12,7 @@
 |9|[Palindrome Number][009]|Math|
 |13|[Roman to Integer][013]|Math, String|
 |14|[Longest Common Prefix][014]|String|
+|19|[Remove Nth Node From End of List][019]|Linked List, Two Pointers|
 
 
 ## Medium
@@ -38,3 +39,4 @@
 [009]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/009/README.md
 [013]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/013/README.md
 [014]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/014/README.md
+[019]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/019/README.md

note/019/README.md

@@ -0,0 +1,65 @@
+# [Remove Nth Node From End of List][title]
+
+## Description
+
+Given a linked list, remove the *n*<sup>th</sup> 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.
+
+**Tags:** Linked List, Two Pointers
+
+## 思路
+
+题意是让你删除链表中的倒数第n个数，我的解法是利用双指针，这两个指针相差n个元素，当后面的指针扫到链表末尾的时候，自然它前面的那个指针所指向的下一个元素就是要删除的元素，即`pre.next = pre.next.next;`，但是如果一开始后面的指针指向的为空，此时代表的意思就是要删除第一个元素，即`head = head.next;`。
+
+``` java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+public class Solution {
+    public ListNode removeNthFromEnd(ListNode head, int n) {
+        ListNode pre = head;
+        ListNode afterPreN = head;
+        while (n-- != 0) {
+            afterPreN = afterPreN.next;
+        }
+        if (afterPreN != null) {
+            while (afterPreN.next != null) {
+                pre = pre.next;
+                afterPreN = afterPreN.next;
+            }
+            pre.next = pre.next.next;
+        } else {
+            head = head.next;
+        }
+        return head;
+    }
+}
+```
+
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我GitHub上的LeetCode题解：[awesome-java-leetcode][ajl]
+
+
+
+[title]: https://leetcode.com/problems/remove-nth-node-from-end-of-list
+[ajl]: https://github.com/Blankj/awesome-java-leetcode

project/LeetCode/leetcode/src/main/java/com/blankj/easy/_019/Solution.java

@@ -14,16 +14,62 @@
 
 public class Solution {
 
-    class ListNode {
+    static class ListNode {
         int val;
         ListNode next;
 
         ListNode(int x) {
             val = x;
         }
+
+        @Override
+        public String toString() {
+            String str = "[" + String.valueOf(val);
+            ListNode p = next;
+            while (p != null) {
+                str += ", " + String.valueOf(p.val);
+                p = p.next;
+            }
+            return str + "]";
+        }
+    }
+
+    public ListNode removeNthFromEnd(ListNode head, int n) {
+        ListNode pre = head;
+        ListNode afterPreN = head;
+        while (n-- != 0) {
+            afterPreN = afterPreN.next;
+        }
+        if (afterPreN != null) {
+            while (afterPreN.next != null) {
+                pre = pre.next;
+                afterPreN = afterPreN.next;
+            }
+            pre.next = pre.next.next;
+        } else {
+            head = head.next;
+        }
+        return head;
     }
 
     public static void main(String[] args) {
         Solution solution = new Solution();
+        ListNode listNode0 = new ListNode(1);
+        ListNode listNode1 = new ListNode(2);
+        ListNode listNode2 = new ListNode(3);
+        ListNode listNode3 = new ListNode(4);
+        ListNode listNode4 = new ListNode(5);
+        listNode0.next = listNode1;
+        listNode1.next = listNode2;
+        listNode2.next = listNode3;
+        listNode3.next = listNode4;
+        listNode4.next = null;
+        System.out.println(listNode0.toString());
+        System.out.println(solution.removeNthFromEnd(listNode0, 2).toString());
+
+        ListNode listNode5 = new ListNode(1);
+        System.out.println(listNode5.toString());
+        ListNode res = solution.removeNthFromEnd(listNode5, 1);
+        System.out.println(res == null ? "[]" : res.toString());
     }
 }