LeetCode——rotate-list

Question

Given a list, rotate the list to the right by k places, where k is non-negative.
For example:
Given1->2->3->4->5->NULLand k =2,
return4->5->1->2->3->NULL.

Solution

1. 典型的fast-slow指针问题。

2. 有很多小细节，k指向最后一个节点，k指向第一个节点，k大于所有节点的个数，链表中节点个数为1，k为0，都要考虑进去。

Code

/**
 * 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 == NULL || k == 0)
            return head;
        int count = 0;
        ListNode* tmp = head;
        while (tmp != NULL) {
        	tmp = tmp->next;
            count++;
        }
        if (k > count) {
            if (k % count == 0)
                return head;
            k = k % count;
        }
        
        ListNode* slow = head;
        ListNode* fast = head;
        int i;
        for (i = 0; i < k - 1; i++) {
            if (fast != NULL) {
                fast = fast->next;
            } else
                return head;
        }
        
        ListNode* pre = NULL;
        while (fast != NULL && fast->next != NULL) {
            pre = slow;
            slow = slow->next;
            fast = fast->next;
        }
        if (pre == NULL)
            return head;
        else
            pre->next = NULL;
        if (fast != NULL)
        	fast->next = head;
        head = slow;
        
        return head;
    }
};