readme

Author: tech-cow

README.md

@@ -11,7 +11,8 @@ Success is like pregnancy, Everybody congratulates you but nobody knows how many
 ## Linked List
 |  #  | Title | Solution | Time | Space | Difficulty |Tag| Note|
 |-----|-------| -------- | ---- | ------|------------|---|-----|
-|234|[Palindrome Linked List](https://leetcode.com/problems/palindrome-linked-list/) | [Python](./linkedlist/palindrome.py) | _O(n)_| _O(1)_  | Easy |CC189| Two Pointers|
 |21| [Merge Two Sorted Lists](https://leetcode.com/problems/merge-two-sorted-lists) | [Python](./linkedlist/MergeTwoSortedLists.py) | _O(n)_| _O(n)_ | Easy || Iteratively|
 |  89  | [Partition List](https://leetcode.com/problems/partition-list/) | [Python](./linkedlist/PartitionLinkedList.py) | _O(n)_ | _O(n)_ | Easy |CC189| |
 |  141  | [Partition List](https://leetcode.com/problems/linked-list-cycle/#/description) | [Python](./linkedlist/LinkedListCycle.py) | _O(n)_ | _O(1)_ | Easy |CC189| Two Pointers|
+|160|[Intersection of Two Linked Lists](https://leetcode.com/problems/intersection-of-two-linked-lists) | [Python](./linkedlist/IntersectionOfTwoLinkedLists.py) | _O(n)_| _O(1)_  | Easy || |
+|234|[Palindrome Linked List](https://leetcode.com/problems/palindrome-linked-list/) | [Python](./linkedlist/palindrome.py) | _O(n)_| _O(1)_  | Easy |CC189| Two Pointers|

linkedlist/IntersectionOfTwoLinkedLists.py

@@ -0,0 +1,32 @@
+class Solution:
+    # @param two ListNodes
+    # @return the intersected ListNode
+    def getIntersectionNode(self, headA, headB):
+        #Count the length for both listnode
+        curA ,curB = headA ,headB
+        len_a = self.lengthcount(curA)
+        len_b = self.lengthcount(curB)
+
+        # Reset pointer, because they both pointers point to the end already.....
+        curA ,curB = headA ,headB
+
+        #pad
+        if len_a > len_b:
+            for _ in xrange(len_a - len_b):
+                curA = curA.next
+        elif len_b > len_a:
+            for _ in xrange(len_b - len_a):
+                curB = curB.next
+
+        while curA != curB:
+            curA = curA.next
+            curB = curB.next
+
+        return curA
+
+    def lengthcount(self, head):
+        count = 0
+        while head:
+            count += 1
+            head = head.next
+        return count