[Leetcode] 61. Rotate List 解题报告

题目：

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：

最正统的解法就是two pointers：定义两个指针p和q，然后先让q前进k步，接着让p和q同步前进直到q到达末尾。此时p的下一个节点就是未来新的头节点，所以此时修改p和q的next即可。本题除了需要注意two pointers的运行边界情况之外，还需要特别注意几个特殊情况：1）链表本身为空；2）k值大于链表的长度；3）更新后的k值为0。本题的时间复杂度为O(n)，空间复杂度为O(1)。

代码：

/**
 * 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) {       // speicial case
            return head;
        }
        ListNode *p = head, *q = head;
        int len = 1;
        while (q->next != NULL) {           // get the length of the list
            len++;
            q = q->next;
        }
        k = k % len;                        // in case k is equal or larger than the length of the list
        if (k == 0) {
            return head;
        }
        q = head;
        for (int i = 0; i < k; ++i) {       // set the second pointer
            q = q->next;
        }
        while (q->next != NULL) {
            p = p->next;
            q = q->next;
        }
        ListNode* newHead = p->next;
        p->next = NULL;
        q->next = head;
        return newHead;
    }
};