一、思路：

? ? ? 先實現兩個節點翻轉，再實現k個節點翻轉，最后實現一組一組翻轉

二、C++代碼：

class Solution { public:ListNode* swap(ListNode *f1, ListNode *f2, int count) {ListNode *nowNode = f2;ListNode *f3, *l1, *l2 = NULL, *l3;while (--count > 0) {if (nowNode->next == NULL) //表明鏈表不足k位return NULL;else if (count == 1) {l2 = nowNode->next;l1 = nowNode;l3 = l2->next;f3 = f2->next;f2->next = l3;if (l1 == f2)l2->next = f2;else {l1->next = f2;l2->next = f3;}if (f1 != NULL)f1->next = l2;return l2;}nowNode = nowNode->next;}return l2;}ListNode *swapKNodes(ListNode *f1, ListNode *f2, int k) {ListNode *nowNode = NULL;ListNode *parent = f2;for (int i = k; i > 1; i -= 2) {if (i == k) {f2 = swap(f1, f2, i);nowNode = f2;}else {nowNode->next = swap(nowNode, nowNode->next, i);nowNode = nowNode->next;}}return f2;}ListNode* reverseKGroup(ListNode* head, int k) {if (k <= 1)return head;ListNode *nowNode = head, *f1 = NULL, *f2 = head, *l2 = NULL;int count = 0;while (nowNode != NULL) {if (count%k == k - 1) {nowNode = f2;if (count == k - 1) {head = f2 = swapKNodes(f1, f2, k);}else {f2 = swapKNodes(f1, f2, k);}f1 = f2;f1 = nowNode;}else if (count%k == 0) {f2 = nowNode;}count++;nowNode = nowNode->next;}return head;} };

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總結

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