update: a difficult link list question

Author: junbinding

list.reverse-nodes-in-k-group.py

@@ -0,0 +1,78 @@
+
+class ListNode:
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+class Solution:
+    """
+    25. K 个一组翻转链表
+    https://leetcode-cn.com/problems/reverse-nodes-in-k-group/
+    给你一个链表，每 k 个节点一组进行翻转，请你返回翻转后的链表。
+    k 是一个正整数，它的值小于或等于链表的长度。
+    如果节点总数不是 k 的整数倍，那么请将最后剩余的节点保持原有顺序。
+    """
+    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
+        dummy = ListNode(0)
+        dummy.next = head
+        pre, end = dummy, dummy
+        while end is not None:
+            i = 0
+            # 循环获取待处理的最后一个节点
+            while i < k and end:
+                end = end.next
+                i += 1
+
+            # 如果没有，则代表最后一轮不足 k 个，不用处理
+            if not end: break
+
+            # 断开链表
+            start = pre.next
+            next = end.next
+            end.next = None
+
+            # 翻转数组
+            pre.next = self.reverse(start)
+            start.next = next
+
+            # 进行下一组处理
+            pre = start
+            end = pre
+        return dummy.next
+
+    def reverse(self, head):
+        pre = None
+        cur = head
+        while cur is not None:
+            tmp = cur.next
+            cur.next = pre
+            pre = cur
+            cur = tmp
+        return pre
+
+    # 转换数组到链表
+    def convertArr2Link(self, arr):
+        if len(arr) == 0:
+            return
+        res = n = ListNode(0)
+        for i in arr:
+            n.next = ListNode(i)
+            n = n.next
+
+        return res.next
+
+
+
+so = Solution()
+# node = so.reverseList()
+params = [1, 2, 3, 4, 5, 6, 7, 8]
+
+# 打印链表
+def printList(link):
+    print(link.val)
+    while(link.next is not None):
+        link = link.next
+        print(link.val)
+
+printList(so.reverse(so.convertArr2Link(params)))
+