9.17 练手腾讯50题 61旋转链表

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本文详细解析了LeetCode第61题“旋转链表”的解决方案，通过寻找截断位置并调整ListNode.next属性实现链表右旋k位。代码采用Python实现，时间复杂度为O(n)。

Leetcode 61. Rotate List

Given a linked list, rotate the list to the right by k places, where k is non-negative.

Example 1:

Input: 1->2->3->4->5->NULL, k = 2
Output: 4->5->1->2->3->NULL
Explanation:
rotate 1 steps to the right: 5->1->2->3->4->NULL
rotate 2 steps to the right: 4->5->1->2->3->NULL

Example 2:

Input: 0->1->2->NULL, k = 4
Output:

2->0->1->NULL

Explanation:
rotate 1 steps to the right: 2->0->1->NULL
rotate 2 steps to the right: 1->2->0->NULL
rotate 3 steps to the right:

0->1->2->NULL

rotate 4 steps to the right:


2->0->1->NULL

想法是直接寻找截断的位置，然后截断更换ListNode.next，代码如下：

class Solution(object):
    def rotateRight(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if(head == None or k == 0):
            return head
        cur = head
        n = 0 #get the length of list
        while(cur):  
            n += 1
            cur = cur.next
        k = k % n
        if(k == 0):
            return head
        secPoint = n-k
        Alist = ListNode(0)
        parent = None
        cur = head
        while(secPoint > 0):
            secPoint -= 1
            parent = cur
            cur = cur.next
        Alist.next = cur
        while(cur.next!=None):
            cur = cur.next 
        cur.next = head
        parent.next = None
        return Alist.next

时间复杂度是O（n）, beat 30%+ 有点低。