From 1aab307ed7a0e9c539bc450fdf7fbdc3082a1b67 Mon Sep 17 00:00:00 2001
From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com>
Date: Sat, 14 Aug 2021 23:18:15 +0800
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用更简单的逻辑完成这道题。
---
 problems/面试题02.07.链表相交.md | 37 ++++++++-----------------
 1 file changed, 12 insertions(+), 25 deletions(-)

diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md
index 9acda71c..8c3a5831 100644
--- a/problems/面试题02.07.链表相交.md
+++ b/problems/面试题02.07.链表相交.md
@@ -160,34 +160,21 @@ Python：
 
 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
+        """
+        根据快慢法则，走的快的一定会追上走得慢的。
+        在这道题里，有的链表短，他走完了就去走另一条链表，我们可以理解为走的快的指针。
 
-        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
+        那么，只要其中一个链表走完了，就去走另一条链表的路。如果有交点，他们最终一定会在同一个
+        位置相遇
+        """
+        cur_a, cur_b = headA, headB     # 用两个指针代替a和b
 
-        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
+        while cur_a != cur_b:
+            cur_a = cur_a.next if cur_a else headB      # 如果a走完了，那么就切换到b走
+            cur_b = cur_b.next if cur_b else headA      # 同理，b走完了就切换到a
+        
+        return cur_a
 ```
 
 Go：