添加 面试题02.07.链表相交 python3版本

Author: morningsky

problems/面试题02.07.链表相交.md

@@ -151,7 +151,44 @@ public class Solution {
 ```
 
 Python：
-
+```python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
+        lengthA,lengthB = 0,0
+        curA,curB = headA,headB
+        while(curA!=None): #求链表A的长度
+            curA = curA.next
+            lengthA +=1
+        
+        while(curB!=None): #求链表B的长度
+            curB = curB.next
+            lengthB +=1
+        
+        curA, curB = headA, headB
+
+        if lengthB>lengthA: #让curA为最长链表的头，lenA为其长度
+            lengthA, lengthB = lengthB, lengthA
+            curA, curB = curB, curA
+
+        gap = lengthA - lengthB #求长度差
+        while(gap!=0): 
+            curA = curA.next #让curA和curB在同一起点上
+            gap -= 1
+        
+        while(curA!=None):
+            if curA == curB:
+                return curA
+            else:
+                curA = curA.next
+                curB = curB.next
+        return None
+```
 
 Go：