Update 092._reverse_linked_list_ii.md

Author: 0xkookoo

docs/Leetcode_Solutions/092._reverse_linked_list_ii.md

@@ -1,7 +1,30 @@
-###92. Reverse Linked List II 
+# 92. Reverse Linked List II
 
-题目:
-<https://leetcode.com/problems/reverse-linked-list-ii/>
+**<font color=red>难度: Medium</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/reverse-linked-list-ii/description/
+
+> 内容描述
+
+```
+Reverse a linked list from position m to n. Do it in one-pass.
+
+Note: 1 ≤ m ≤ n ≤ length of list.
+
+Example:
+
+Input: 1->2->3->4->5->NULL, m = 2, n = 4
+Output: 1->4->3->2->5->NULL
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(N)******- 空间复杂度: O(1)******
 
 
 难度:
@@ -14,7 +37,7 @@ Medium
 
 AC 代码
 
-```
+```python
 class Solution(object):
     def reverseBetween(self, head, m, n):
         """
@@ -23,44 +46,66 @@ class Solution(object):
         :type n: int
         :rtype: ListNode
         """
-        # m == n, not reverse
-        if m == n : return head
-
-        dummy = ListNode(-1)
+        if not head or m == n: 
+            return head
+        node_m_before = dummy = ListNode(-1)
         dummy.next = head
-
-        mbefore = dummy
-        cnt = 1
 
-        while mbefore and cnt < m:
-            mbefore = mbefore.next
-            cnt += 1
+        for i in range(m-1):
+            node_m_before = node_m_before.next
 
         prev = None
-        cur = mbefore.next
-        tail1 = mbefore.next
+        cur = node_m_before.next
+        node_m = node_m_before.next
 
-
-        while cnt <= n :
+        for i in range(n-m+1):
             nxt = cur.next
             cur.next = prev
             prev = cur
             cur = nxt
-            cnt += 1
 
-
-
-        mbefore.next = prev
-        tail1.next = cur
+        node_m_before.next = prev
+        node_m.next = cur
 
         return dummy.next 
 ```
 
+> 思路 2
+******- 时间复杂度: O(N)******- 空间复杂度: O(1)******
+
 看了一下别人的代码，又比我写的好嘛，因为是保证m和n有效，用的是for循环先找到 m node:
 
+详细见[realisking](https://leetcode.com/problems/reverse-linked-list-ii/discuss/30709/Talk-is-cheap-show-me-the-code-(and-DRAWING))
+```
+for _ in range(m-1):
+	....
 
-	for _ in range(m-1):
-		....
+for _ in range(n-m):
+	reverse 操作
 		
-	for _ in range(n-m):
-		reverse 操作
+```
+```python
+class Solution(object):
+    def reverseBetween(self, head, m, n):
+        """
+        :type head: ListNode
+        :type m: int
+        :type n: int
+        :rtype: ListNode
+        """
+        if not head or m == n: 
+            return head
+        p = dummy = ListNode(None)
+        dummy.next = head
+        for i in range(m-1): 
+            p = p.next
+        tail = p.next
+
+        for i in range(n-m):
+            tmp = p.next                  
+            p.next = tail.next            
+            tail.next = tail.next.next    
+            p.next.next = tmp            
+        return dummy.next
+```
+