题解 | #链表中的节点每k个一组翻转#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    ListNode* reverseKGroup(ListNode* head, int k) {
        // write code here
        ListNode *res = new ListNode(0);
        res->next = head;
        ListNode *pre = res;
        ListNode *curr = head;
        int cnt = 0;
        //获取链表总长度
        while(head)
        {
            ++cnt;
            head = head->next;
        }
        int n = cnt/k;//需要翻转的区间数 = 总长度/区间长度
        for(int i = 0;i<n;++i)
        {
            //翻转k个结点时，需要进行k-1次操作
            for(int j = 1;j<k;++j)
            {
                //头插法翻转该区间
                ListNode *temp = curr->next;
                curr->next = temp->next;
                temp->next = pre->next;
                pre->next = temp;
            }
            pre = curr;
            curr = curr->next;
        }
        return res->next;
    }
};