LeetCode OJ--Rotate List

http://oj.leetcode.com/problems/rotate-list/

取得后面k个节点，然后截断插到前面。如果k比list长，则按照求余算。

去后面的k个节点：使用两个指针，第一个指针比第二个指针先走k步，然后两个一起往后走，等到第一个到达最后一个节点，第二个就是倒数第k个节点。

#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) {
        //找到倒数第k个节点
        ListNode *pEnd = head,*pLastK = head;
        int _k = k;
        if(k==0 ||head == NULL)
            return head;

        //计算共有多少个节点
        ListNode *node_count = head;
        int count = 0;
        while(node_count)
        {
            node_count = node_count->next;
            count++;
        }
        if(_k>count)
            _k = k%count;
        if(_k == 0)
            return head;

        while(_k--)
        {
            if(pEnd == NULL)
                return head;
            pEnd = pEnd->next;
        }
        if(pEnd == NULL)
            return head;
        while(pEnd->next)
        {
            pEnd = pEnd->next;
            pLastK = pLastK->next;
        }
        //截断
        ListNode *ans = pLastK->next;
        pLastK->next = NULL;
        //再插到前面
        ListNode *temp = ans;
        while(temp->next)
        {
            temp = temp->next;
        }
        temp->next = head;
        return ans;
    }
};

int main()
{
    Solution  myS;
    ListNode *n1 = new ListNode(1);
    ListNode *n2 = new ListNode(2);
    ListNode *n3 = new ListNode(3);
    ListNode *n4 = new ListNode(4);
    ListNode *n5 = new ListNode(5);
    n1->next = n2;
    n2->next = n3;
    n3->next = n4;
    n4->next = n5;
    myS.rotateRight(n4,4);
    return 0;
}