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Author: assassint2017

Reverse Nodes in k-Group/Reverse_Nodes_in_k-Group.cpp

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+// 其实并不难，就是比较麻烦，如果非要说技巧，应该是单链表翻转的时候如何把头结点的next指针置空
+// Runtime: 16 ms, faster than 96.91% of C++ online submissions for Reverse Nodes in k-Group.
+// Memory Usage: 9.9 MB, less than 35.48% of C++ online submissions for Reverse Nodes in k-Group.
+
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+
+class Solution 
+{
+public:
+    ListNode* reverseKGroup(ListNode* head, int k) 
+    {   
+        // 首先统计一下链表的长度
+        int length = 0;
+        ListNode* temp = head;
+        while (temp)
+        {
+            ++length;
+            temp = temp->next;
+        }
+        
+        // 边界条件处理
+        if (head == nullptr || k <= 0 || k > length) return head;
+        
+        ListNode dummyHead(666);
+        ListNode* preHead = &dummyHead;
+        
+        int numOfLeft = length;
+        ListNode *preNode = nullptr, *curNode = head, *nextNode = nullptr;
+        
+        while (numOfLeft)
+        {
+            if (numOfLeft < k) break;
+            
+            ListNode *curHead = curNode, *curTail = nullptr;
+            
+            for (int i = 0; i < k; ++i)
+            {   
+                nextNode = curNode->next;
+                curNode->next = preNode;
+                
+                preNode = curNode;
+                curNode = nextNode;
+                
+                --numOfLeft;
+            }
+
+            curTail = preNode;
+            
+            preHead->next = curTail;
+            curHead->next = curNode;
+            
+            preHead = curHead;
+        }
+        
+        return dummyHead.next;
+    }
+};