[LeetCode] 061. Rotate List

• Problem (Medium)
• Approach 1: (My Solution - Beat 65.38%)
  • Idea
  • Solution
  • Complexity

Problem (Medium)

061. 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

Approach 1: (My Solution - Beat 65.38%)

Idea

1. Connect the last item in the linked list to the head;
2. Walk through the cycle linked list to the n-k-1-th item;
3. Save n-k-th item as res;
4. Set the n-k-1-th item’s next to `None;
5. Return res.

Solution

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

class Solution1:
    def rotateRight(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if not head:
            return
        tmp = head
        count = 0
        while tmp:
            count += 1
            if not tmp.next:
                tmp.next = head
                break
            else:
                tmp = tmp.next
        tmp = head
        for _ in range(count - k % count - 1):
            tmp = tmp.next
        res = tmp.next
        tmp.next = None
        return res

Complexity

• Time: $O(n)$
• Space: $O(1)$

KF