Reverse Nodes in k-Group
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这道题目,我的解法思想与Swap Nodes in Pairs那道题基本上是一样的,只不过在翻转group内节点的时候,我使用了数组栈。我的代码如下:
ListNode* reverseKGroup(ListNode* head, int k) { int cnt = 0; // 计数 ListNode *sstack[k], *cursor; // 数组栈 ListNode dummy(0), *pre = &dummy, *p = head; // p为待翻转group的第一个节点 pre->next = head; // pre为p之前的节点 while (p) { cnt = k; cursor = p; for (int i=0; i<k; i++) { if (cursor == NULL) { return dummy.next; } sstack[i] = cursor; // group内节点入栈 cursor = cursor->next; } for (int i=k-1; i>=1; i--) { // 翻转group内各节点 sstack[i]->next = sstack[i-1]; } pre->next = sstack[k-1]; // 一些连接操作 p->next = cursor; pre = p; p = p->next; } return dummy.next; }
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