LeetCode 061. Rotate List

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.

這道題要求是將鏈表最後K個節點移到鏈表開頭，其實挺簡單的。

但是需要注意一點：當K大於鏈表長度的時候，必須從頭開始遍歷。

/**
 * 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) 
    {
        if(!head || k < 0)
            return NULL;
        if(k == 0)
            return head;
        ListNode * rear = head;
        ListNode * fore = head;
        ListNode * temp;
        int i = 0;
        while(i<k)
        {
            fore = fore->next;
            if(!fore)   //k的長度大於鏈表長度
                fore = head;
            i++;
        }
        while(fore->next)
        {
            fore = fore->next;
            rear = rear->next;
        }
        //此時rear指向待旋轉節點的前一個節點
        //fore指向最後一個節點
        fore->next = head;
        head = rear->next;
        rear->next = NULL;
        return head;
    }
};

因爲while循環裏面忘了寫i++，硬是不知道哪兒出錯了，最後在VS裏面才調試出來，囧死！

測試代碼：

int main()
{
	int num[2]={1,2};
	struct ListNode *head = new ListNode(0);
	struct ListNode *temp = head;
	for(int i=0; i<2; i++)
	{
		struct ListNode * p = new ListNode(num[i]);
		temp ->next = p;
		temp = p;
	}
	temp = head;
	head = head->next;
	delete temp;

	cout<<"原始鏈表：";
	temp = head;
	while(temp)
	{
		cout<<temp->val<<" ";
		temp = temp->next;
	}
	cout<<endl;

	Solution sol;
	head = sol.rotateRight(head, 1);

	cout<<"操作後鏈表：";
	while(head)
	{
		cout<<head->val<<" ";
		temp = head;
		head = head->next;
		delete temp;
	}
	cout<<endl;
	return 0;
}