leetcode 25 Reverse Nodes in k-Group

最新推荐文章于 2025-05-21 19:43:17 发布

ghoshorn

最新推荐文章于 2025-05-21 19:43:17 发布

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本文介绍了一种算法，该算法将链表中的节点每K个一组进行反转，并返回修改后的链表。讨论了特殊情况处理及实现细节。

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Reverse Nodes in k-Group

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

类似leetcode 24的Swap Nodes in Pairs

每次先看是否存在k个节点，如果够的话再逆转。

头k个的逆转和之后的逆转分开。

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    # @param head, a ListNode
    # @param k, an integer
    # @return a ListNode
    def reverseKGroup(self, head, k):
        if head==None:
            return None
        if k==1:
            return head
        cnt=1
        p=head
        while p and cnt<k:
            p=p.next
            cnt+=1
        if cnt<k or p==None:
            return head
        p1=head
        p2=p1.next
        p3=p2.next
        cnt=1
        p1.next=p.next
        pre=p1
        while cnt<k:
            p2.next=p1
            p1=p2
            p2=p3
            if p3:
                p3=p3.next
            cnt+=1
        head=p

        while  True:
            cnt=1
            p=pre.next
            while p and cnt<k:
                p=p.next
                cnt+=1
            if cnt<k or p==None:
                break
            p1=pre.next
            p2=p1.next
            p3=p2.next
            cnt=1
            while cnt<k:
                p2.next=p1
                p1=p2
                p2=p3
                <span style="color:#ff0000;">if p3:</span>
                    p3=p3.next
                cnt+=1
            tmp=pre.next
            # pre.next=p2
            <span style="color:#ff0000;">pre.next=p1 #mod</span>
            pre=tmp
            <span style="color:#ff0000;">tmp.next=p2 #add </span>

        return head

if __name__ == '__main__':
    head=ListNode(1)
    head.next=ListNode(2)
    head.next.next=ListNode(3)
    head.next.next.next=ListNode(4)
    # head.next.next.next.next=ListNode(5)
    p=head
    while p:
        print p.val,
        p=p.next
    print
    a=Solution()
    p=a.reverseKGroup(head,2)
    while p:
        print p.val,
        p=p.next