[LeetCode] 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.

思路一：首先将链表链接成循环链表,并得到链表长度count。计算k%n,再从头开始前进 count - k就是断开位置。

　　　　时间复杂度O(n),空间复杂度O(1)

/**
 * 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;
        ListNode *pter = head;
        int count = 1;
        while (pter->next != NULL) {
            pter = pter->next;
            ++count;
        }
        pter->next = head;
        pter = pter->next;
        k %= count;
        for (int i = 1; i < count - k; ++i) { //找到旋转后头节点的前一个节点
            pter = pter->next;
        }
        head = pter->next;
        pter->next = NULL;
        
        return head;
    }
};

思路二：使用两个指针

/**
 * 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;
        
        ListNode *pfast = head;
        for (int i = 0; i < k; ++i) { 
            if (pfast->next != NULL) {
                pfast = pfast->next;
            } else {
                pfast = head;
            }
        }
        
        ListNode *pslow = head; //指向新的头节点的前一个节点
        while (pfast->next != NULL) {
            pfast = pfast->next;
            pslow = pslow->next;
        }
        pfast->next = head;
        head = pslow->next;
        pslow->next = NULL;
        
        return head;
    }
};