LC.61 | 旋转链表 | 链表 | 双指针拆分链表

输入：链表

要求：将链表每个节点向右移动 k 个位置

输出：移动后的链表

思路：先定位新head位置，然后双指针拆分后重组

/**
 * 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) {
        ListNode* dummy = new ListNode(0, head);
        ListNode* h = dummy->next;
        
        int len = 0; 
        while (h) {
            len++;
            h = h->next;
        }
        if (len == 0) {
            return head;
        }
        k = k % len;
        if (k == 0) {
            return head;
        }

        k = len - k;
        ListNode* t1 = dummy;
        ListNode* t2;

        while (k--) {
            t1 = t1->next;
        }
        
        ListNode* ans = new ListNode(0, t1->next);
        t2 = t1->next;
        t1->next = NULL;
        while (t2->next) {
            t2 = t2->next;
        }
        t2->next = dummy->next;
        return ans->next;
    }
};