019_Remove_Nth_Node_From_End_of_List

Author: qiyuangong

README.md

@@ -23,6 +23,7 @@ Also, there are open source implementations for basic data structs and algorithm
 | 15 | [3Sum](https://leetcode.com/problems/3sum/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/015_3Sum.py) | 1. Sort and O(n^2) search with three points <br>2. Multiple Two Sum (Problem 1) |
 | 16 | [3Sum Closest](https://leetcode.com/problems/3sum-closest/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/016_3Sum_Closest.py) | Sort and Multiple Two Sum check abs|
 | 18 | [4Sum](https://leetcode.com/problems/4sum/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/018_4Sum.py) | The same as 3Sum, but we can merge pairs with the same sum |
+| 19 | [Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/019_Remove_Nth_Node_From_End_of_List.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/019_Remove_Nth_Node_From_End_of_List.java) | 1. Go through list and get length, then remove length-n, O(n) and O(n)<br>2. Two pointers, first pointer goes to n position, then move both pointers until reach tail, O(n) and O(n) |
 | 20 | [Valid Parentheses](https://leetcode.com/problems/valid-parentheses/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/020_Valid_Parentheses.py) | 1. Stack<br>2. Replace all parentheses with '', if empty then True |
 | 21 | [Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/021_Merge_Two_Sorted_Lists.py) | Add a dummy head, then merge two sorted list in O(m+n) |
 | 23 | [Merge k Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/023_Merge_k_Sorted_Lists.py) | 1. Priority queue O(nk log k)<br>2. Binary merge O(nk log k)|  |

java/019_Remove_Nth_Node_From_End_of_List.java

@@ -0,0 +1,63 @@
+/*Given a linked list, remove the n-th node from the end of list and return its head.
+
+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.
+
+Follow up:
+Could you do this in one pass?*/
+
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    /*public ListNode removeNthFromEnd(ListNode head, int n) {
+        ListNode curr = head;
+        int ls = 0;
+        while (curr != null) {
+            curr = curr.next;
+            ls++;
+        }
+        // n == len
+        if (ls == n) {
+            if (ls > 1) return head.next;
+            return null;
+        }
+        curr = head;
+        // Move to ls - n - 1
+        for (int i = 0; i < ls - n - 1; i++) {
+            curr = curr.next;
+        }
+        // Remove ls - n - 1
+        curr.next = curr.next.next;
+        return head;
+    }*/
+
+    public ListNode removeNthFromEnd(ListNode head, int n) {
+        ListNode slow, fast, curr;
+        slow = head; fast = head;
+        for (int i = 0; i < n; i++)
+            fast = fast.next;
+        // n == len
+        if (fast == null) {
+            head = head.next;
+            return head;
+        }
+        // Move both pointers, until reach tail
+        while (fast.next != null) {
+            fast = fast.next;
+            slow = slow.next;
+        }
+        curr = slow.next;
+        slow.next = curr.next;
+        return head;
+    }
+}

python/019_Remove_Nth_Node_From_End_of_List.py

@@ -1,3 +1,15 @@
+# Given a linked list, remove the n-th node from the end of list and return its head.
+#
+# 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.
+#
+# Follow up:
+# Could you do this in one pass?
+
 # Definition for singly-linked list.
 # class ListNode(object):
 #     def __init__(self, x):
@@ -30,7 +42,6 @@ class Solution(object):
     #         index[index_pos].next = index[index_pos + 1].next
     #         return head
 
-
     def removeNthFromEnd(self, head, n):
         # https://leetcode.com/discuss/86721/o-n-solution-in-java
         if head is None: