leetcode25. Reverse Nodes in k-Group

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本文解析了LeetCode上的第25题——按组翻转链表问题，并提供了完整的C++代码实现。该解决方案使用了辅助节点来简化边界条件处理，并通过迭代方式实现了每k个一组的链表节点翻转。

leetcode25. Reverse Nodes in k-Group

class Solution {
public:
    ListNode* reverseKGroup(ListNode* head,int k) {
        int step = k;
        ListNode *front = new ListNode(0);
        front->next = head;
        ListNode *tail = front;
        vector<ListNode*> heads = {};
        ListNode *lHead = front;
        if(head==NULL){
            return NULL;
        }
        if(k==0 || k==1) return head;
        tail = tail->next;
        while(tail!=NULL){
            front->next= reverseListK(tail,k);
            front = tail;
            tail = tail->next;
        }
        return lHead->next;


    }
    ListNode* reverseListK(ListNode* &head,int k){
        if(head==NULL){
            return NULL;
        }
        ListNode *temp = head;
        int j = k;

        ListNode *h=NULL;
        ListNode *cursor = head;
        ListNode *tail = head;

        while(temp!=NULL&&j>=1){
            cursor = temp;
            temp = temp->next;
            j--;
        }
        if(j>=1){
            h = head;
            head = cursor;
            return h;
        }
        cursor = head;
        while(k>0 && head!=NULL){
            head = head->next;
            cursor->next = h;
            h = cursor;
            cursor = head;
            k--;


        }
        if(k==0) tail->next = cursor;
        head = tail;

        return h;
    }

};