标签：style blog io color 使用 for sp div on
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)
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *rotateRight(ListNode *head, int k) { 12 if (head == NULL || k <= 0) return head; 13 ListNode *pter = head; 14 int count = 1; 15 while (pter->next != NULL) { 16 pter = pter->next; 17 ++count; 18 } 19 pter->next = head; 20 pter = pter->next; 21 k %= count; 22 for (int i = 1; i < count - k; ++i) { //找到旋转后头节点的前一个节点 23 pter = pter->next; 24 } 25 head = pter->next; 26 pter->next = NULL; 27 28 return head; 29 } 30 };
思路二：使用两个指针
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *rotateRight(ListNode *head, int k) { 12 if (head == NULL || k <= 0) return head; 13 14 ListNode *pfast = head; 15 for (int i = 0; i < k; ++i) { 16 if (pfast->next != NULL) { 17 pfast = pfast->next; 18 } else { 19 pfast = head; 20 } 21 } 22 23 ListNode *pslow = head; //指向新的头节点的前一个节点 24 while (pfast->next != NULL) { 25 pfast = pfast->next; 26 pslow = pslow->next; 27 } 28 pfast->next = head; 29 head = pslow->next; 30 pslow->next = NULL; 31 32 return head; 33 } 34 };
标签：style blog io color 使用 for sp div on
原文地址：http://www.cnblogs.com/vincently/p/4059943.html
