61. Rotate List

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

My Solutions:

• 找到list的末尾5，同时计算list的长度

• 把5和list现在的head1，连成环

• 用len-k算出需要移的node的前一个。比如如果右移2步，则5 - 2 = 3 之后的nodes会到前面来

• 找到3，先把3之后的node存成新的head，再把3之后的node变成null

• 返回新的head，这个head的后面已经和之前的头部相连

• Time: O(n); Space: O(1)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if (head == null || head.next == null || k == 0) return head;
        
        int len = 1;
        ListNode node = head;
        // 找到list的末尾node
        while (node.next != null) {
            node = node.next;
            len++;
        }
        
        if (k % len == 0) return head; 
        k = k % len; //可能要右移超过1圈
        node.next = head; //最后一个node连上head，变成环
        
        // 找到需要分割的地方
        for (int i = 1; i < len - k; i++) {
            head = head.next;
        } //此时head = 3
        
        ListNode res = head.next; //res = 4
        head.next = null; // 3的之后是null
        
        return res;
    }
}