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
https://www.nowcoder.com/practice/b49c3dc907814e9bbfa8437c251b028e
/** * 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(head->next == nullptr || head == nullptr || k < 2) { return head; } int size = 0; ListNode *count = head; while(count != nullptr) { size++; count = count->next; } int group = size / k; ListNode *pre = nullptr; ListNode *cur = head; ListNode *next = nullptr; ListNode *tail = head; ListNode *nHead = nullptr; ListNode *nTail = nullptr; ListNode *nNext = nullptr; for(int i=1;i<=k*group;i++) { tail = tail->next; } for(int i=1;i<=k*group;i++) { next = cur->next; cur->next = pre; pre = cur; cur = next; } nTail = tail; for(int i=1;i<=k*group;i++) { if(i % k == 1) { nHead = pre; } nNext = pre->next; if(i % k == 0) { pre->next = nTail; nTail = nHead; } pre = nNext; } return nTail; } };