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

Solution (Java)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if (head == null || k == 0) {
            return head;
        }
        ListNode tail = head;
        // find the count and let tail points to last node
        int count = 1;
        while (tail != null && tail.next != null) {
            count++;
            tail = tail.next;
        }
        // calculate number of times to rotate by count modulas
        int times = k % count;
        if (times == 0) {
            return head;
        }
        ListNode temp = head;
        // iterate and go to the K+1 th node from the end or count - K - 1 node from
        // start
        for (int i = 1; i <= count - times - 1 && temp != null; i++) {
            temp = temp.next;
        }
        ListNode newHead = null;
        if (temp != null && tail != null) {
            newHead = temp.next;
            temp.next = null;
            tail.next = head;
        }
        return newHead;
    }
}

Solution (Javascript)

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} k
 * @return {ListNode}
 */
var rotateRight = function(head, k) {
    var count = 1;
    var last = head;
    var now = head;

    if (!head || !head.next) return head;

    while (last.next) {
        last = last.next;
        count++;
    }

    k %= count;

    if (k === 0) return head;

    while (k < count - 1) {
        now = now.next;
        k++;
    }

    last.next = head;
    head = now.next;
    now.next = null;

    return head;
};

Explain:

1. 拿到长度 count 和最后一个 last
2. k %= count
3. 找到新的最后一个 newLast，last.next = head，head = newLast.next，newLast.next = null
4. return head

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).