yxc题解 借鉴方法（常用！）

class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (!head) return head;
        int n = 0;
        for (auto p = head; p; p = p->next) n ++;
        k %= n;
        if (!k) return head;
        auto tail = head;
        while (tail->next) tail = tail->next;
        auto p = head;
        for (int i = 1; i <= n - k - 1; i ++) p = p->next;
        tail->next = head;
        head = p->next;
        p->next = NULL;
        return head; 
    }
};

这样写也可以

class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (!head) return head;
        int n = 0;
        auto tail = head;
        for (; tail->next; tail = tail->next) n ++; // 这里是tail->next!=NULL，因为tail = tail->next会执行使最终tail指向链表最后一个节点
        n++;

        k %= n;
        if (!k) return head;
        // auto tail = head;
        // while (tail->next) tail = tail->next;
        auto p = head;
        for (int i = 1; i <= n - k - 1; i ++) 
            p = p->next;
        tail->next = head; //但是将这行放在n++下面就不行了
        head = p->next;
        p->next = NULL;
        return head;
    }
};

另一种写法：(借鉴y总题解写法)

class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (!head) return head;
        int n = 0;
        for (auto p = head; p; p = p->next) n ++;
        auto first = head, second = head;
        k %= n;
        if (!k) return head;

        for (int i = 1; i <= k; i ++) 
            first = first->next;
        int t = n - k - 1;
        while (t --) {
            first = first->next;
            second = second->next;
        }

        first->next = head;
        head = second->next;
        second->next = 0;
        return head;
    }
};