0061._Rotate_List.md

61. Rotate List

难度:Medium

刷题内容

原题连接

• https://leetcode.com/problems/rotate-list/

内容描述

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

思路1

- 时间复杂度: O(n)- 空间复杂度: O(n)******

这里的n是指链表的长度，我们用k对n的长度取余，因为k大于链表的长度之后就是一个循环。不需要重复操作。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        int len = 0;
        ListNode* h = head;
        ListNode* pre;
        while(h)
        {
            ++len;
            pre = h;
            h = h ->next;
        }
        if(!len)
            return head;    
        int count1 = len - k % len;
        if(count1 == len)
            return head;
        h = head;
        ListNode* tail;
        while(count1)
        {
            tail = h;
            h = h ->next;
            --count1;
        }
        pre ->next = head;
        tail ->next = nullptr;
        return h;
    }
};