leetCode61. 旋转链表
题目思路：见如图所示

代码展示

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if(!head) return head;

        int n; // 找出链表长度
        ListNode* tail; // 找到尾结点
        for(auto p = head; p; p = p->next){
            tail = p;
            n++;
        }

        // k的值可能大于n,我们要去取有效的
        k %= n;
        if(k == 0) return head;

        auto p = head;
        for(int i = 0; i < n - k - 1; i++) p = p->next;

        tail->next = head;
        head = p->next;
        p->next = nullptr;

        return head;
    }
};