Swap Nodes in Pairs: Accepted

README.md

@@ -138,6 +138,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Reverse Linked List](problems/src/linked_list/ReverseLinkedList.java) (Easy)
 - [Delete Node in a Linked List](problems/src/linked_list/DeleteNode.java) (Easy)
 - [Reverse Nodes in k-Group](problems/src/linked_list/ReverseNodesKGroup.java) (Hard)
+- [Swap Nodes in Pairs](problems/src/linked_list/SwapNodesInPairs.java) (Medium)
 
 #### [Math](problems/src/math)
 

problems/src/linked_list/SwapNodesInPairs.java

@@ -0,0 +1,58 @@
+package linked_list;
+
+/**
+ * Created by gouthamvidyapradhan on 13/08/2017.
+ * Given a linked list, swap every two adjacent nodes and return its head.
+
+ For example,
+ Given 1->2->3->4, you should return the list as 2->1->4->3.
+
+ Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.
+ */
+public class SwapNodesInPairs {
+
+    public static class ListNode{
+        int val;
+        ListNode next;
+        ListNode(int x){
+            val = x;
+        }
+    }
+
+    public static void main(String[] args) throws Exception{
+        ListNode node = new ListNode(1);
+        node.next = new ListNode(2);
+        node.next.next = new ListNode(3);
+        node.next.next.next = new ListNode(4);
+        node.next.next.next.next = new ListNode(5);
+        node.next.next.next.next.next = new ListNode(6);
+        ListNode head = new SwapNodesInPairs().swapPairs(node);
+        while(head != null){
+            System.out.println(head.val);
+            head = head.next;
+        }
+    }
+
+    public ListNode swapPairs(ListNode head) {
+        if(head == null || head.next == null)
+            return head;
+        ListNode newHead = head.next;
+        ListNode curr = head.next;
+        ListNode prev = head;
+        ListNode prevPrev = new ListNode(-1); //dummy node
+        while(curr != null){
+            prev.next = curr.next;
+            curr.next = prev;
+            prevPrev.next = curr;
+            if(prev.next != null){
+                curr = prev.next.next;
+                prev = prev.next;
+                prevPrev = prevPrev.next.next;
+            } else {
+                curr = null;
+            }
+        }
+        return newHead;
+    }
+
+}