From dae5f4debe408c2fce12226a5e7b986d116df094 Mon Sep 17 00:00:00 2001
From: "Zhen Liu (Ramsey)" <416034116@qq.com>
Date: Mon, 17 May 2021 01:18:28 -0300
Subject: [PATCH] =?UTF-8?q?Update=20=E9=9D=A2=E8=AF=95=E9=A2=9802.07.?=
 =?UTF-8?q?=E9=93=BE=E8=A1=A8=E7=9B=B8=E4=BA=A4=20Java=20=E7=89=88?=
 =?UTF-8?q?=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

代码经过测试 beats 98% submission, 完全follow Carl's C++版本的全部逻辑，稍作修改并加上了注释！辛苦啦！谢谢
---
 problems/面试题02.07.链表相交.md | 58 ++++++++++++++++++++++++-
 1 file changed, 57 insertions(+), 1 deletion(-)

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 97045ed2..493438d7 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
@@ -92,7 +92,63 @@ public:
 
 
 Java：
-
+```Java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) {
+ *         val = x;
+ *         next = null;
+ *     }
+ * }
+ */
+public class Solution {
+    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
+        ListNode curA = headA;
+        ListNode curB = headB;
+        int lenA = 0, lenB = 0;
+        while (curA != null) { // 求链表A的长度
+            lenA++;
+            curA = curA.next;
+        }
+        while (curB != null) { // 求链表B的长度
+            lenB++;
+            curB = curB.next;
+        }
+        curA = headA;
+        curB = headB;
+        // 让curA为最长链表的头，lenA为其长度
+        if (lenB > lenA) {
+            //1. swap (lenA, lenB);
+            int tmpLen = lenA;
+            lenA = lenB;
+            lenB = tmpLen;
+            //2. swap (curA, curB);
+            ListNode tmpNode = curA;
+            curA = curB;
+            curB = tmpNode;
+        }
+        // 求长度差
+        int gap = lenA - lenB;
+        // 让curA和curB在同一起点上（末尾位置对齐）
+        while (gap-- > 0) {
+            curA = curA.next;
+        }
+        // 遍历curA 和 curB，遇到相同则直接返回
+        while (curA != null) {
+            if (curA == curB) {
+                return curA;
+            }
+            curA = curA.next;
+            curB = curB.next;
+        }
+        return null;
+    }
+    
+}
+```
 
 Python：