Rotate List

自己搞定的，刚开始题目还没弄清

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

For example:
Given 1->2->3->4->5->NULL and k = 2,

return 4->5->1->2->3->NULL.

分析： (关键在于k可能会很大)

1. list 是空的吗？

2. 连成一个圈，再k 取余， 再转

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param head, a ListNode
    # @param k, an integer
    # @return a ListNode
    def rotateRight(self, head, k):
        if head==None:
            return head
            
        cur = head
        count = 1
        while cur.next!=None:
            cur = cur.next
            count += 1
        cur.next = head
        
        k = k % count
        cur = head
        for dummy_i in range(count-k-1):
            cur = cur.next
        newhead = cur.next
        cur.next = None
        
        return newhead
        

C++

/**
 * 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){
            return head;
        }
        
        ListNode *cur = head;
        int count = 1;
        while (cur->next!=NULL){
            cur = cur->next;
            count += 1;
        }
        cur->next = head;
        
        k = k % count;
        cur = head;
        for(i=1;i<=(count-k-1);i++){
            cur = cur->next;
        }
        ListNode *newhead = cur->next;
        cur->next = NULL;
        
        return newhead;
        
    }
};