leetcode25. Reverse Nodes in k-Group

Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list.

k is a positive integer and is less than or equal to the length of the linked 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 list’s nodes, only nodes themselves may be changed.
Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]
Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]
给你链表的头节点 head ，每 k 个节点一组进行翻转，请你返回修改后的链表。

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。

你不能只是单纯的改变节点内部的值，而是需要实际进行节点交换。

class Solution:
    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        h = t = ListNode(0, head)
        p, stack = head, []        
        while True:
            for _ in range(k):
                if p: 
                    stack.append(p)     # 压栈
                    p = p.next
                else: return h.next     # 不满栈，直接返回
            for _ in range(k):
                t.next = stack.pop()    # 出栈，同时续到链表尾部
                t = t.next
            t.next = p                  # 连接后段链表

原文地址：https://blog.csdn.net/weixin_44245188/article/details/143662356

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