LeetCode Reverse Nodes in k-Group

LeetCode解题之Reverse Nodes in k-Group

---

原题

将一个链表中每k个数进行翻转，末尾不足k个的数不做变化。

注意点：

• 不允许修改节点的值
• 只能用常量的额外空间

例子：

输入: head = 1->2->3->4->5, k = 2
输出: 2->1->4->3->5

输入: head = 1->2->3->4->5, k = 3
输出: 3->2->1->4->5

解题思路

这个题是Swap Nodes in Pairs的升级版。我们来看一下翻转k个节点要进行哪些操作，A->B->C->D->E，现在我们要翻转BCD三个节点。进行以下几步：

1. C->B
2. D->C
3. B->E
4. A->D
5. 返回及节点B

上面做了两件事，把k个节点先翻转（1、2两步），再和前后两个节点连接起来（3、4两步）。

AC源码

# Definition for singly-linked list.class ListNode(object):    def __init__(self, x):        self.val = x        self.next = Noneclass Solution(object):    def reverseKGroup(self, head, k):        """        :type head: ListNode        :type k: int        :rtype: ListNode        """        if not head or k <= 1:            return head        dummy = ListNode(-1)        dummy.next = head        temp = dummy        while temp:            temp = self.reverseNextK(temp, k)        return dummy.next    def reverseNextK(self, head, k):        # Check if there are k nodes left        temp = head        for i in range(k):            if not temp.next:                return None            temp = temp.next        # The last node when the k nodes reversed        node = head.next        prev = head        curr = head.next        # Reverse k nodes        for i in range(k):            nextNode = curr.nex            curr.next = prev            prev = curr            curr = nextNode        # Connect with head and tail        node.next = curr        head.next = prev        return node

欢迎查看我的Github (https://github.com/gavinfish/LeetCode-Python) 来获得相关源码。