[数据结构与算法][LeetCode] 61. 旋转链表

给你一个链表的头节点 head ，旋转链表，将链表每个节点向右移动 k 个位置。

示例 1：

输入：head = [1,2,3,4,5], k = 2
输出：[4,5,1,2,3]

?

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/rotate-list

/**
 * 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 || !head->next)
            return head;
        ListNode* fast = head;
        ListNode* slow = head;
        int list_length = 0;

        while(fast){
            list_length ++;
            fast = fast->next;
        }
        k = k % list_length;
        fast = head;
        while(k--)
            fast = fast->next;

        while(fast->next){
            slow = slow->next;
            fast = fast->next;
        }
        fast->next = head;
        head = slow->next;
        slow->next = nullptr;

        return head;
    }
};

官方解法：

class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (k == 0 || head == nullptr || head->next == nullptr) {
            return head;
        }
        int n = 1;
        ListNode* iter = head;
        while (iter->next != nullptr) {
            iter = iter->next;
            n++;
        }
        int add = n - k % n;
        if (add == n) {
            return head;
        }
        iter->next = head;
        while (add--) {
            iter = iter->next;
        }
        ListNode* ret = iter->next;
        iter->next = nullptr;
        return ret;
    }
};

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/rotate-list/solution/xuan-zhuan-lian-biao-by-leetcode-solutio-woq1/
来源：力扣（LeetCode）
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