LeetCode 61: Rotate List (链表好题)

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本文解析如何使用O(n)时间复杂度解决链表旋转问题，通过节点计数和连接首尾巧妙简化操作。理解链表旋转的原理并学习高效算法实现。

61. Rotate List

Medium

Given the head of a linked list, rotate the list to the right by k places.

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [4,5,1,2,3]

Example 2:

Input: head = [0,1,2], k = 4
Output: [2,0,1]

Constraints:

The number of nodes in the list is in the range [0, 500].
-100 <= Node.val <= 100
0 <= k <= 2 * 109

解法1：

先将链表首尾相连，然后从尾往后移动n-k步就到了新链表的前一个位置，断开链表并返回新链表的位置即可。

/**
 * 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 == nullptr) return nullptr;
        ListNode *p = head;
        int cnt = 1;
        while (p->next) {
            p = p->next;
            cnt++;
        }
        k %= cnt; 
        p->next = head;
        
        for (int i = 0; i < cnt - k; i++) p = p->next;
        head = p->next;
        p->next = nullptr;
        return head;        
    }
};