Add 'Reverse Nodes in k-Group' solution.

Author: liuyingshan

Reverse_Nodes_in_kGroup .cpp

@@ -0,0 +1,63 @@
+/*
+  Given a linked list, reverse the nodes of a linked list k at a time and
+  return its modified list.
+  If the number of nodes is not a multiple of k then left-out nodes in the
+  end should remain as it is.
+  You may not alter the values in the nodes, only nodes itself may be
+  changed.
+
+  Only constant memory is allowed.
+
+  For example,
+  Given this linked list: 1->2->3->4->5
+  For k = 2, you should return: 2->1->4->3->5
+  For k = 3, you should return: 3->2->1->4->5
+*/ 
+
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    // Reverse sub-linked list which starts from head node.
+    // preHead's next node is the head node of the reversed sub-linked list.
+	// The function returns the tail node of the reversed sub-linked list.
+    ListNode *subReverse(ListNode *preHead, ListNode *head, int cnt) {
+        if (cnt == 0) {
+            preHead->next = head;
+        } else {
+            subReverse(preHead, head->next, cnt - 1)->next = head;
+        }
+        return head;
+    }
+    
+    ListNode *reverseKGroup(ListNode *head, int k) {
+        if (head == NULL || k <= 1) return head;
+        
+        ListNode *newHead = new ListNode(-1);
+        newHead->next = head;
+        ListNode *st = newHead, *en, *tail;
+        while (st != NULL) {
+        	// Check whether last nodes is enough to be reversed. 
+            int cnt = 0;
+            en = st->next;
+            while (cnt < k && en != NULL) {
+                ++cnt;
+                en = en->next;
+            }
+            
+            if (cnt < k) break;
+            tail = subReverse(st, st->next, k-1);
+            tail->next = en;
+            st = tail;
+        }
+        head = newHead->next;
+        delete newHead;
+        return head;
+    }
+};