(链表) leetcode 25. Reverse Nodes in k-Group

解法一：用栈来实现链表反转

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        // O(k) 空间复杂度
        if(!head)
            return NULL;
        ListNode dummy(0);
        stack<ListNode*> st;
        dummy.next = head;
        ListNode* cur = &dummy, *next = head;
        while(next){
            for(int i=0; i<k && next; ++i){
                st.push(next);
                next = next->next;
            }
            if(st.size() != k)
                return dummy.next;  //剩下的元素<k个
            while(st.size()){
                cur->next = st.top();
                st.pop();
                cur = cur->next;
            }
            cur->next = next;
        }
        return dummy.next;
    }
};

解法二：找到需要反转的区间，依次将后一个反转到前面.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        //k为需要reverse的个数
        //无辅助空间
        if(head == NULL)
            return NULL;
        ListNode dummy(0);
        dummy.next = head;
        ListNode* pre = &dummy;
        while(pre){
            pre = reverse(pre, k);
        }
        return dummy.next;
    }
    
    ListNode* reverse(ListNode* pre, int k){
        ListNode* last = pre;   //pre始终指向区间的前一个结点
        for(int i=0; i<=k; i++){
            last = last->next;  //last指向k区间的下一个元素
            if(i!=k && last==NULL)
                //区间内不够k个
                return NULL;  //返回
        }
        ListNode* tail = pre->next;
        ListNode* cur = tail->next;
        while(cur!=last){
            ListNode* next = cur->next;
            cur->next = pre->next;
            pre->next = cur;
            tail->next = next;
            cur = next;
        }
        return tail;
    }
};