LeetCode61 旋转链表

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
思路：定位四个节点：1，3，4，5（例子1中），分别对应两段的开头和结尾

public ListNode rotateRight(ListNode head, int k) {
        if(head == null || head.next == null) return head;
        ListNode fast = head;
        ListNode slow = head;
        ListNode pre = new ListNode(0);
        pre.next = head;
        int s = 0;
        while(fast.next != null){
            fast = fast.next;
            s++;
        }
        k = k % (s+1);
        if(k == 0) return head;
        for(int i = 0; i < s - k + 1; i++){
            slow = slow.next;
            pre = pre.next;
        }
        fast.next = head;
        pre.next = null;
        return slow;
    }