[LeetCode]Rotate List-优快云博客

本文链接：https://blog.youkuaiyun.com/sunbaigui/article/details/8981272

本文介绍了一种链表右旋算法的具体实现方法，通过双指针技术有效地完成了链表节点的旋转操作。该算法首先确定链表长度，并计算实际需要旋转的位置，然后使用两个指针分别指向链表的不同位置，最终完成链表的旋转。

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struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
//draw some pictures and then think it twice, just be careful when process pointers
public:
	ListNode *rotateRight(ListNode *head, int k) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		ListNode* run = head;
		int len = 0;
		for(; run != NULL; run = run->next, len++);
		if(len == 0) return NULL;
		k = k%len;
		if(k == 0) return head;

		ListNode* second = head;
		ListNode* first = head;
		while (k--)
			first = first->next;
		while (first->next != NULL)
		{
			second = second->next;
			first = first->next;
		}
		ListNode* newHead = second->next;
		second->next = NULL;
		first->next = head;
		return newHead;
	}
};

second time

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(head == NULL) return head;
        int len = 0;
        ListNode* tmp = head;
        while(tmp != NULL) len++, tmp = tmp->next;
        k = k%len;
        if(k == 0) return head;
        
        ListNode* fast = head;
        ListNode* slow = head;
        int i;
        for(i = 0; i < k && fast != NULL; ++i) fast = fast->next;
        //if(i != k-1)//...k is too large k > len?
        while(fast != NULL)
        {
            fast = fast->next;
            slow = slow->next;
        }
        //if(slow == head) return head;
        ListNode* newHead = slow;
        ListNode* oldSlow = slow;
        while(slow->next != NULL) slow = slow->next;
        slow->next = head;
        while(slow->next != oldSlow) slow = slow->next;
        slow->next = NULL;
        
        return newHead;
    }
};