Reverse Nodes in K-Groups Leetcode Python

标签：python   leetcode   linklist   

For k = 3, you should return: 3->2->1->4->5

这题做法与reverse linklist和 reverse pair类似都需要有一个用来reverse的函数。

题目的意思是每k个node翻转一次，如果不够k个就不转。

# 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 reverse(self,start,end):
        dummy=ListNode(0)
        dummy.next=start
        while dummy.next!=end:
            tmp=start.next
            start.next=tmp.next
            tmp.next=dummy.next
            dummy.next=tmp
        return end,start
        
    def reverseKGroup(self, head, k):
        if head==None:return None
        dummy=ListNode(0)
        dummy.next=head
        start=dummy
        while start.next:
            end=start
            for i in range(k-1):
                end=end.next
                if end.next==None:
                    return dummy.next
            newstart,newend=self.reverse(start.next,end.next)
            start.next=newstart
            start=newend
        return dummy.next
            
    
        

Reverse Nodes in K-Groups Leetcode Python

标签：python   leetcode   linklist   

原文地址：http://blog.csdn.net/hyperbolechi/article/details/43082921