61. Rotate List #
Problem #
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
Summary #
Rotate the linked list K times.
Solution Approach #
The point to note in this problem is that K may be very large, K = 2000000000, so if you use a loop it will definitely time out. You should find an algorithm with O(n) complexity. Since it is a cyclic rotation, the final state is actually determined; using the length of the linked list to take the remainder can obtain the final rotated result of the linked list.
This problem also cannot be solved with recursion; a recursive solution will time out.
Code #
package leetcode
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func rotateRight(head *ListNode, k int) *ListNode {
if head == nil || head.Next == nil || k == 0 {
return head
}
newHead := &ListNode{Val: 0, Next: head}
len := 0
cur := newHead
for cur.Next != nil {
len++
cur = cur.Next
}
if (k % len) == 0 {
return head
}
cur.Next = head
cur = newHead
for i := len - k%len; i > 0; i-- {
cur = cur.Next
}
res := &ListNode{Val: 0, Next: cur.Next}
cur.Next = nil
return res.Next
}