思路:
先将链表闭合成环;
再找到相应的位置断开这个环,确定新的链表头和链表尾。
程序很简单:
#include<iostream>
using namespace std;
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 || head->next == NULL || k <= 0)
return head;
ListNode*tail = head;
int n = 1;
while (tail->next != NULL)
{
tail = tail->next;
n++;
}
tail->next = head;
//start
if (k >= n)
k = k % n;
int temp = n - k - 1;
while (temp--)
{
head = head->next;
}
ListNode*newhead = head->next;
head->next = NULL;
return newhead;
}
};
int main()
{
ListNode*node1 = new ListNode(1);
ListNode*node2 = new ListNode(2);
ListNode*node3 = new ListNode(3);
ListNode*node4 = new ListNode(4);
ListNode*node5 = new ListNode(5);
node1->next = node2; node2->next = node3; node3->next = node4; node4->next = node5;
Solution s;
ListNode*newhead = s.rotateRight(node1, 2);
ListNode*temp = newhead;
while (temp != NULL)
{
cout << temp->val << " ";
temp = temp->next;
}
system("pause");
return 0;
}