35：Reverse Nodes in k-Group

最新推荐文章于 2022-04-08 22:28:12 发布

weishenmetlc

最新推荐文章于 2022-04-08 22:28:12 发布

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分类专栏： leetcode

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本文介绍了一种算法，该算法可以每k个一组反转链表中的节点，并返回修改后的链表。文章详细展示了如何实现这一功能的代码，包括如何处理不是k的倍数的节点数量以及如何仅使用常量内存进行操作。

题目：Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
You may not alter the values in the nodes, only nodes itself may be changed.
Only constant memory is allowed.
For example, Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5

解题代码如下：

class Solution {
public:
        ListNode* reverseKGroup(ListNode* head, int k) {
                ListNode dummy{-1};
                dummy.next = head;
                ListNode *prev = &dummy, *cur = &dummy;
                // cur 指向第 k 个节点，prev 指向第一个节点之前的节点
                for (int i = 0; cur && i != k; ++i)
                        cur = cur -> next;
                if (!cur) return head;

                for (int i = 0; ; ++i) {
                        // 每隔 k 个节点就颠倒这 k 个节点的顺序
                        if (i % k == 0) {
                                ListNode* tmp = prev -> next;
                                reverseList(prev -> next, cur -> next);
                                prev -> next = cur;
                                cur = tmp;
                        }
                        if (!cur -> next) break;
                        prev = prev -> next;
                        cur = cur -> next;
                }
                return dummy.next;
        }

private:
        //颠倒[first, last)内的节点的顺序
        ListNode* reverseList(ListNode* first, ListNode* last) {
                ListNode dummy{-1};
                dummy.next = first;
                while (first -> next != last) {
                        ListNode* tmp = first -> next;
                        first -> next = tmp -> next;
                        tmp -> next = dummy.next;
                        dummy.next = tmp;
                }
                return dummy.next;
        }
};