Merge pull request #275 from betNevS/master

添加 02.07.链表相交 go版本

problems/0142.环形链表II.md

@@ -234,25 +234,19 @@ class Solution:
 ```
 
 Go：
-```func detectCycle(head *ListNode) *ListNode {
+```go
-    if head ==nil{
+func detectCycle(head *ListNode) *ListNode {
+    slow, fast := head, head
+    for fast != nil && fast.Next != nil {
+        slow = slow.Next
+        fast = fast.Next.Next
+        if slow == fast {
+            for slow != head {
+                slow = slow.Next
+                head = head.Next
+            }
             return head
         }
-    slow:=head
-    fast:=head.Next
-
-    for fast!=nil&&fast.Next!=nil{
-        if fast==slow{
-            slow=head
-            fast=fast.Next
-            for fast!=slow {
-                fast=fast.Next
-                slow=slow.Next
-            }
-            return slow
-        }
-        fast=fast.Next.Next
-        slow=slow.Next
     }
     return nil
 }

problems/面试题02.07.链表相交.md

@@ -155,6 +155,42 @@ Python：
 
 Go：
 
+```go
+func getIntersectionNode(headA, headB *ListNode) *ListNode {
+    curA := headA
+    curB := headB
+    lenA, lenB := 0, 0
+    // 求A，B的长度
+    for curA != nil {
+        curA = curA.Next
+        lenA++
+    }
+    for curB != nil {
+        curB = curB.Next
+        lenB++
+    }
+    var step int
+    var fast, slow *ListNode
+    // 请求长度差，并且让更长的链表先走相差的长度
+    if lenA > lenB {
+        step = lenA - lenB
+        fast, slow = headA, headB
+    } else {
+        step = lenB - lenA
+        fast, slow = headB, headA
+    }
+    for i:=0; i < step; i++ {
+        fast = fast.Next
+    }
+    // 遍历两个链表遇到相同则跳出遍历
+    for fast != slow {
+        fast = fast.Next
+        slow = slow.Next
+    }
+    return fast
+}
+```
+
 javaScript: 
 
 ```js