61.Rotate List

Tags: Medium Link List

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

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Given a linked list, rotate the list to the right by k places, where k is non-negative.

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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

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Answer:

/**
 * 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) {
        ListNode *dummy = new ListNode(-1);
        dummy -> next = head;
        ListNode *cur, *pre;
        pre = dummy;
        int num=0; //用来统计链表元素个数

        while(pre -> next) {++num; pre = pre -> next;}

        if(num == 0 || num == 1) return head;

        for(int i = 0; i < k % num; ++i)
        {
            pre = dummy;
            cur = pre -> next;

            while(cur -> next) {cur = cur -> next;pre = pre -> next;}
            cur -> next = dummy -> next;
            pre -> next = NULL;
            dummy -> next = cur;
        }

        return dummy -> next;
    }
};

此种方法速度最快：Runtime: 8 ms, faster than 100.00% of C++ online submissions for Rotate List.Memory Usage: 8.9 MB, less than 100.00% of C++ online submissions for Rotate List.

Answer2:

/**
 * 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) {
        if(!head || k == 0) return head;

        ListNode *p = head;
        int len=1;

        while(p -> next) //求链表的长度
        {
            ++len;
            p = p -> next;
        }
        k = len - k % len;

        p -> next = head; //首尾相连
        for(int step = 0; step < k; ++step)
        {
            p = p -> next;
        }
        head = p -> next; //新的首节点
        p -> next = NULL; //断开环

        return head;
    }
};

此种方法是先首尾相连形成一个环，最后断开环，速度一般：Runtime: 16 ms, faster than 40.50% of C++ online submissions for Rotate List.