LeetCode之“链表”：Rotate List

　　题目链接

　　题目要求：

　　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.

　　比较直观的一个解法就是先求出链表长度，然后再确定从头节点位移到分割节点处的步数。接下来就是分割链表和拼接了。具体程序（8ms）如下：

/**
 * 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 || !head->next || k == 0)
            return head;
            
        int len = 0;
        ListNode *start = head;
        while(start)
        {
            len++;
            start = start->next;
        }
        k = len - k % len;
        if(k == 0 || k == len)
            return head;
            
        ListNode *dummy = new(nothrow) ListNode(INT_MIN);
        assert(dummy);
        
        start = head;
        for(int i = 0; i < k - 1; i++)
            start = start->next;
        
        dummy->next = start->next;
        start->next = nullptr;
        
        start = dummy;
        while(start->next)
            start = start->next;
        start->next = head;
        
        head = dummy->next;
        delete dummy;
        dummy = nullptr;
        
        return head;
    }
};

　　这题还有另一个很巧妙的解法：首先从head开始跑，直到最后一个节点，这时可以得出链表长度len。然后将尾指针指向头指针，将整个圈连起来，接着往前跑len – k%len，从这里断开，就是要求的结果了。

　　具体程序（12ms）如下：

/**
 * 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 || !head->next || k == 0)
            return head;
            
        int len = 1;
        ListNode *start = head;
        while(start->next)
        {
            len++;
            start = start->next;
        }
        k = len - k % len;
        if(k == 0 || k == len)
            return head;
        
        start->next = head;
        for(int i = 0; i < k; i++)
            start = start->next;
        
        head = start->next;
        start->next = nullptr;

        return head;
    }
};