lintcode:Rotate List

最新推荐文章于 2024-08-12 20:30:19 发布

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本文介绍了一种算法，用于将单链表向右旋转k个位置。通过双指针技巧实现高效旋转，避免了传统方法中可能存在的多次反转操作。文章详细解释了算法的设计思路，并提供了完整的C++代码实现。

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Given a list, rotate the list to the right by k places, where k is non-negative.

Example

Given 1->2->3->4->5->null and k=2
return 4->5->1->2->3->null

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    /**
     * @param head: the list
     * @param k: rotate to the right k places
     * @return: the list after rotation
     */
    int ListCount(ListNode* head)
    {
        if (!head)
            return 0;
        
        int count = 1;
        
        ListNode *pCurNode = head;
        
        while (pCurNode->next)
        {
            pCurNode = pCurNode->next;
            count++;
        }
        
        return count;
    }
     
    ListNode *rotateRight(ListNode *head, int k) {
        // write your code here
        if (!head)
            return NULL;
        if (k == 0)
            return head;
        
        int nodeCount =  ListCount(head);
        if (nodeCount == 1)
            return head;
        
        
        ListNode* pFastNode = head;
        ListNode* pSlowNode = head;
        
        ListNode fakeHead = ListNode(-1);
        ListNode* pPreNode  = &fakeHead;
        pPreNode->next = head;
        
        k = k%ListCount(head);
        
        for (int i=1; i<=k-1; i++)
        {
            if (pFastNode->next)
            {
                pFastNode = pFastNode->next;
            }
            else
            {
                return head;
            }
        }
        
        while (pFastNode->next != NULL)
        {
            pFastNode = pFastNode->next;
            pSlowNode = pSlowNode->next;
            pPreNode  = pPreNode->next;
        }
        
        pFastNode->next = head;
        pPreNode->next = NULL;
        
        return pSlowNode;
    }
};