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

AC代码

int calcLen(ListNode* head) {
    int len = 0;
    while (head) {
        len++;
        head = head->next;
    }
    return len;
}

class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if(!head) return NULL;
        int len = calcLen(head), r = k % len;
        ListNode *pre = head, *post = head;
        for (int i = 0; i < r; ++i) post = post->next;
        while (post->next) {
            pre = pre->next;
            post = post->next;
        }
        ListNode* thead = pre->next;
        if (!thead) return head;
        post->next = head;
        pre->next = NULL;
        return thead;
    }
};

总结

求链表长度和实际旋转长度=>找到旋转后的头=>拼接原链表首尾=>切断新头节点和上一个节点的关联