【链表】节点的插入删除【25. Reverse Nodes in k-Group】
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题目链接:https://leetcode.com/problems/reverse-nodes-in-k-group/#/description
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: // -1 -> 2 -> 1 -> 3 -> 4 // pre cur nxt t // 将nxt节点插入在pre和2中间 ListNode* reverseKGroup(ListNode* head, int k) { if(head==NULL||k==1) return head; // 将head链表改为带头节点的链表 ListNode* tmp = new ListNode(-1); tmp->next=head; // 参数:pre:前驱,cur:当前节点,nxt:后继 ListNode* pre=tmp,*cur=tmp,*nxt; // 统计head链表中节点个数 int num=0; while(cur=cur->next) num++; // 做num/k次移位循环 while(k<=num){ cur=pre->next; nxt=cur->next; // 每一段数据做k-1次插入 for(int i=1;i<k;i++){ cur->next=nxt->next; nxt->next=pre->next; pre->next=nxt; nxt=cur->next; } num-=k; pre=cur; } // 返回不带头节点的链表 return tmp->next; }};
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