【LeetCode】 Rotate List

最新推荐文章于 2024-10-08 11:15:15 发布

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本文介绍了一种链表向右循环移位k次的算法实现。通过计算链表长度并利用尾节点连接头节点的方式形成闭环，再通过断开指定位置节点完成循环移位，最终返回新的链表头部。

Rotate List

Total Accepted: 13773 Total Submissions: 62790 My Submissions

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.

这是向右循环移动的意思。要读懂题目的意思

/**
 * 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 == NULL || head->next == NULL)
            return head;
        if (k <= 0) return head;
        int len = 0;
        ListNode *p = head, *tail = head;
        while (p) {
            len++;
            tail = p;
            p = p->next;
        }
        
        k = k % len;
        if (k == 0) return head;
        int i = 0;
        p = head;
        while (p) {
            i++;
            if (i == len - k)
                break;
            p = p->next;
        }
        ListNode *tmp_head = p->next;
        p->next = NULL;
        tail->next = head;
        return tmp_head;
    }
};