Rotate List

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.

思路：

这个题目的陷阱中，输入为空指针大约不算其中之一。主要的问题是k可以大于这个链表的总长度。这样，真正的旋转次数是k % length(ListNode)。那么新的head就是length(ListNode) - k  + 1的结点。

题解：

 

/** * 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 == nullptr)            return nullptr;                // first update the length of the list        ListNode* len_iter = head;        int len = 1;        while(len_iter->next != nullptr)            ++len, len_iter = len_iter->next;        k = k % len;                if (k == 0)            return head;                    len_iter->next = head;  // make it a circled link list        ListNode* breakpoint = head;        for(int i = 0; i < len - k - 1; ++i)            breakpoint = breakpoint->next;                head = breakpoint->next;        breakpoint->next = nullptr;                return head;    }};