Reverse Nodes in k-Group

NeetCode/reverse-nodes-in-k-group/step1.py

@@ -0,0 +1,78 @@
+"""
+Problem:
+    separate list and each length is k
+    reverse each segment
+Question:
+Discussion:
+    define function to reverse segment
+    def reverse_segment(head, length)
+
+kthを取ってくることで、dummy.nextを正しく設定できている
+"""
+def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
+    def reverse_list(head):
+        cur = head
+        prev = None
+        count = 0
+        while count < k:
+            tmp = cur.next
+            cur.next = prev
+            prev = cur
+            cur = tmp
+            count += 1
+        return prev, cur
+
+    length = 0
+    cur = head
+    while cur:
+        length += 1
+        cur = cur.next
+    count = 0
+    prev_group_final = ListNode(next=head)
+    group_final = head
+    while length - count >= k:
+        tmp, group_next = reverse_list(group_final)
+        group_final.next = group_next
+        group_final = group_next
+        prev_group_final.next = tmp
+        count += k
+    return dummy.next
+"""
+これだと、dummy.nextが設定できない。reverseされたgroupのheadを持っていないから
+dummy自体のポジションも動かしにくい。
+"""
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
+        def get_kth(head):
+            count = 0
+            while head and count < k:
+                head = head.next
+                count += 1
+            return head
+
+        dummy = ListNode(next=head)
+        group_prev = dummy
+        while True:
+            kth = get_kth(group_prev)
+            if not kth:
+                break
+            group_next = kth.next
+
+            prev = group_next
+            cur = group_prev.next
+            while cur != group_next:
+                tmp = cur.next
+                cur.next = prev
+                prev = cur
+                cur = tmp
+
+            tmp = group_prev.next
+            group_prev.next = kth
+            group_prev = tmp
+        return dummy.next            

NeetCode/reverse-nodes-in-k-group/step3.py

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+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
+        def get_kth(head):
+            count = 0
+            while head and count < k:
+                head = head.next
+                count += 1
+            return head
+
+        dummy = ListNode(next=head)
+        group_prev = dummy
+        while True:
+            kth = get_kth(group_prev)
+            if not kth:
+                break
+            group_next = kth.next
+
+            cur = group_prev.next
+            prev = group_next
+            while cur != group_next:
+                tmp = cur.next
+                cur.next = prev
+                prev = cur
+                cur = tmp
+
+            tmp = group_prev.next
+            group_prev.next = prev
+            group_prev = tmp
+        return dummy.next