170. Rotate List【medium】

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

 

Example

Given 1->2->3->4->5 and k = 2, return 4->5->1->2->3.

解法一：

public class Solution {
    private int getLength(ListNode head) {
        int length = 0;
        while (head != null) {
            length ++;
            head = head.next;
        }
        return length;
    }
    
    public ListNode rotateRight(ListNode head, int n) {
        if (head == null) {
            return null;
        }
        
        int length = getLength(head);
        n = n % length;
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        
        ListNode tail = dummy;
        for (int i = 0; i < n; i++) {
            head = head.next;
        }
        
        while (head.next != null) {
            tail = tail.next;
            head = head.next;
        }
        
        head.next = dummy.next;
        dummy.next = tail.next;
        tail.next = null;
        return dummy.next;
    }
}

快慢指针  + dummy节点，参考@NineChapter 的代码

解法二：

class Solution {
public:
    /**
     * @param head: the list
     * @param k: rotate to the right k places
     * @return: the list after rotation
     */
    ListNode *rotateRight(ListNode *head, int k) {
        if (head == NULL) {
            return head;
        }
        
        int len = 0;
        for (ListNode *node = head; node != NULL; node = node->next) {
            len++;
        }
        k = k % len;
        
        if (k == 0) {
            return head;
        }
        
        ListNode *fast = head;
        for (int i = 0; i < k; i++) {
            fast = fast->next;
        }
        
        ListNode *slow = head;
        while (fast->next != NULL) {
            slow = slow->next;
            fast = fast->next;
        }
        
        fast->next = head;
        head = slow->next;
        slow->next = NULL;
        
        return head;
    }
};

不带dummy节点的解法，参考@NineChapter 的代码