原题描述

旋转链表

思路

简单题，可使用模拟法。注意处理空链表、k >=链表长度len等边界情况

/**
 * 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* cur = head;
        if (cur == nullptr) return 0;
        int len = 0;
        while (cur) ++len, cur = cur->next;
        int first = (len - (k % len)) % len;
        if (first == 0) return head;

        cur = head;
        for (int i = 0; i < first - 1; ++i) {
            cur = cur->next;
        }
        ListNode* last = cur->next;
        cur->next = nullptr;

        ListNode* ans = last;
        while (last->next) last = last->next;
        last->next = head;

        return ans;
    }
};

运行结果

执行用时：8 ms, 击败82.17%的用户
内存消耗：11.4 MB，超过44.80%的用户