题解 | #链表中的节点每k个一组翻转#
链表中的节点每k个一组翻转
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/** * struct ListNode { * int val; * struct ListNode *next; * ListNode(int x) : val(x), next(nullptr) {} * }; */ class Solution { public: /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param head ListNode类 * @param k int整型 * @return ListNode类 */ ListNode* reverseKGroup(ListNode* head, int k) { if(k==1) return head; ListNode* dummy = new ListNode(-1); dummy->next = head; ListNode* pre = dummy; ListNode* cur = dummy->next; int coun = 0; while(cur){ coun ++; cur = cur->next; } cur = dummy->next; int cou = coun/k; while( cou-- ){ int n = k-1; while(n--){ ListNode* tmp = cur->next; cur->next = tmp->next; tmp->next = pre->next; pre->next = tmp; } pre = cur; cur = cur->next; } return dummy->next; } };#题解分享#