Merge pull request #830 from dsrao711/issue_740

Issue 740

Author: Saviour1001

DSA 450 GFG/next_greater_element.py

@@ -0,0 +1,34 @@
+#https://leetcode.com/problems/reverse-linked-list/
+
+# Iterative method
+#Approach : 
+
+# Store the head in a temp variable called current . 
+
+#   curr = head , prev = null 
+
+# Now for a normal linked list , the current will point to the next node and so on till null 
+# For reverse linked list, the current node should point to the previous node and the first node here will point to null 
+
+# Keep iterating the linkedlist until the last node and keep changing the next of the current node to prev node and also 
+# update the prev node to current node and current node to next node
+
+
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+
+class Solution:
+    
+    def reverseList(self, head):
+        curr = head
+        prev = None
+        while(curr != None):
+          next = curr.next
+          curr.next = prev
+          prev = curr
+          curr = next
+          
+        return prev
+          

DSA 450 GFG/reverse_linked_list_iterative.py

@@ -0,0 +1,23 @@
+# Approach :
+# Divide the linked list to two halved
+# First half is head and the remaining as rest 
+# The head points to the rest in a normal linked list 
+# In the reverse linked list , the next of current points to the prev node and the head node should point to NULL
+# Keep continuing this process till the last node 
+
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+
+
+class Solution:
+    def reverseList(self, head):
+        if head is None or head.next is None:
+          return head
+        rest = self.reverseList(head.next)
+        head.next.next = head
+        head.next = None
+        return rest