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README.md

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

Related Topics:
Linked List, Two Pointers

Similar Questions:

Solution 1.

// OJ: https://leetcode.com/problems/rotate-list/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
    int getLength(ListNode *head) {
        int len = 0;
        for (; head; head = head->next, ++len);
        return len;
    }
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (!head) return NULL;
        int len = getLength(head);
        k %= len;
        if (k == 0) return head;
        auto p = head, q = head;
        while (k--) q = q->next;
        while (q->next) {
            p = p->next;
            q = q->next;
        }
        auto h = p->next;
        q->next = head;
        p->next = NULL;
        return h;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/rotate-list/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (!head) return NULL;
        int len = 1;
        auto p = head;
        for (; p->next; ++len, p = p->next);
        p->next = head;
        k = len - k % len;
        while (k--) p = p->next;
        head = p->next;
        p->next = NULL;
        return head;
    }
};