[LeetCode][python3]0025. Reverse Nodes in k-Group

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N2I -2020.03.19

1. My first solution

class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        if not head or k<2:
            return head
        seq=[0]
        p1=head
        index=1
        while p1!=None:
            seq.append(p1)
            p1=p1.next
            if index%k==0 and index!=0:
                #print("swap",index-k+1,index+1)
                seq[index-k+1:index+1]=seq[index:index-k:-1]
            index+=1
        
        h=p1=ListNode(0)
        for i in seq[1:]:
            p1.next=i
            p1=p1.next
            #print(i.val)
        p1.next=None
        return h.next

N2I -2020.03.19

Note:

• seq[6:3:-1] Reverse order of seq[4:7], where seq[6] is included and seq[3] is not.

print("example of s[a:b:-1]")
s=[0,1,2,3,4,5]
print(s[1:4])
    #>>[1,2,3]
print(s[3:0:-1])
    #>>[3,2,1]
print(s[3:0:-2])
    #>>[3,1]
print(s[0:4])
    #>>[0,1,2,3]
print(s[3::-1])
    #>>[3,2,1,0]
print(s[4:-1:-1]) #but -1 still represent len(s) here
    #>>[]

Explanation:

Same way to solve it with Q.24 and others.