[LeetCode] Rotate Linked List

最新推荐文章于 2025-09-06 17:22:57 发布

jiyanfeng1

最新推荐文章于 2025-09-06 17:22:57 发布

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分类专栏：链表文章标签： leetcode

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本文介绍了一种链表右旋算法的实现方法，通过计算链表长度和旋转位置，将链表从指定位置断开并重新连接，形成新的链表结构。

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.

Analysis:
The idea is to get the whole length of the list, then compute the rotate position.
And cut the list from the rotate position, link the front part to the last part.

e.g.
1->2->3->4->5->null , k =2;
len = 5;
rotate pos = 5-2 = 3;

cut list: 1->2->3 ---> 4->5->null
link : 4->5->1->2->3->null

    ListNode *rotateRight(ListNode *head, int k) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head==NULL || k==0) {return head;}
         
        ListNode* p = head;
        int len=0;
        //get total length 
        while (p!=NULL){
            len++;
            p=p->next;
        }
        //get rotate position
        int r;
        if (k%len==0){
            return head;
        }else{
            r = len-(k%len)-1;    
        }
        p=head;
        while (r>0){
            p=p->next;
            r--;
        }
         
         
        //cut current list, link the back to the front
        ListNode *q =p->next;
        if (q==NULL){return head;}
        while (q->next!=NULL){
            q=q->next;
        }
        q->next = head;
        q=p->next;
        p->next=NULL;
         
        return q;
    }