【力扣】K个一组翻转链表

K个一组翻转链表

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* reverseKGroup(ListNode* head, int k) {//某些情况不用翻转if (k == 1 || head->next == nullptr) return head;//计算有多少个节点int n = 0;//n个节点ListNode* begin = head;while (begin){++n;begin = begin->next;}//循环m次，每次进行长度为k的逆序ListNode* guard = new ListNode(0);ListNode* tmp = guard;//作为返回的节点ListNode* cur = head;ListNode* tail = guard;int m = n / k;//循环m次for (int i = 0; i < m; ++i){tail = cur;for (int j = 0; j < k; ++j){ListNode* next = cur->next;cur->next = tmp->next;tmp->next = cur;cur = next;}//需要找到尾节点作为起始位置tmp = tail;} //不需要逆序的节点尾插tail->next = cur;//进行释放指针tmp = guard->next;delete guard;return tmp;}
};