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[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)
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 };
[LeetCode] Rotate List
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