[LeetCode]91. Rotate List旋转链表

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

2

4->5->1->2->3->NULL

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/**
* 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) {
ListNode* curr = head;
int length = 0;
while (curr != NULL) {
++length;
curr = curr->next;
}
if(length > 0) k %= length;
if(length == 0 || length == 1 || k == 0) return head;
head = reverseList(head);
if(k > 1) head = reverseBetween(head, 1, k);
if(k + 1 < length) head = reverseBetween(head, k + 1, length);
return head;
}
private:
ListNode* reverseList(ListNode* head) {
ListNode* rHead = NULL;
ListNode* curr = head;
ListNode* pTail = NULL;
while (curr != NULL) {
ListNode* pNext = curr->next;
if (pNext == NULL)
rHead = curr;
curr->next = pTail;
pTail = curr;
curr = pNext;
}
return rHead;
}
ListNode* reverseBetween(ListNode* head, int m, int n) {
if (head == NULL || head->next == NULL || m == n) return head;
ListNode* help = new ListNode(0);
help->next = head;
ListNode* prev = help;
for (int i = 0; i < m - 1; ++i)
prev = prev->next;
ListNode *curr = prev->next, *next = curr->next, *temp = curr, *last = NULL;
for (int i = m; i < n; ++i) {
last = next->next;
next->next = curr;
curr = next;
next = last;
}
prev->next = curr;
temp->next = last;
return help->next;
}
};

解法2：因为在k处旋转链表只需要断开第n-k个节点和第n-k+1个节点（n为链表长度）然后新链表的头设为第n-k+1个节点，第n-k个节点的next指针设为NULL即可，因此可以先扫描一次链表统计节点个数，然后再从头开始前移n-k-1步，处理好这个节点的next指针即可。

/**
* 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) {
ListNode* curr = head;
int length = 0;
while (curr != NULL) {
++length;
curr = curr->next;
}
if(length > 0) k %= length;
if (length == 0 || length == 1 || k == 0) return head;
curr = head;
for (int i = 0; i < length - k - 1; ++i)
curr = curr->next;
ListNode* newHead = curr->next; // 旋转后新的头节点
curr->next = NULL; // 断开链表
curr = newHead;
while (curr->next != NULL)
curr = curr->next;
curr->next = head; // 原链表的尾节点和头节点链接起来
return newHead;
}
};

转载于:https://www.cnblogs.com/aprilcheny/p/4972501.html