New Simple Approachable LinkedList Solution

Author: rahulcode22

Bit-Manipulation/Power-of-2.py

@@ -0,0 +1,3 @@
+def isPowerofTwo(n):
+    return n and (not(n and (n-1)))
+print isPowerofTwo(9)

Bit-Manipulation/Xor-upto-n.py

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+def xorUptoN(n):
+    if n%4 == 0:
+        return n
+    if n%4 == 1:
+        return 1
+    if n%4 == 2:
+        return n + 1
+    else:
+        return 0
+
+print xorUptoN(6)

LinkedList/Swap-List-nodes-in-pairs.py

@@ -0,0 +1,29 @@
+'''
+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.
+'''
+#Approach 1
+def swapPairs(self, head):
+    pre, pre.next = self, head
+    while pre.next and pre.next.next:
+        a = pre.next
+        b = a.next
+        pre.next, b.next, a.next = b, a, b.next
+        pre = a
+    return self.next
+
+
+#Approach 2
+def swapPairs(head):
+    curr = head
+    nxt = head.next
+    while curr and nxt:
+        curr.val, nxt.val = nxt.val, curr.val
+        curr = nxt.next
+        if curr:
+            nxt = curr.next
+    return head