LeetCode-Reverse Nodes in k-Group
Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5
思路:一开始是想要动态规划的方式,即写一个反转函数,每K个字符调用一次,但是较大的问题就是要记录反转后链表的首尾指针。后来查到可以直接原地反转的方式,如下图
代码如下:
/** * 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) { int cnt = 0; bool isHead = true; //是否链头 ListNode *p = head, *q = NULL, *prev = NULL; while(p){ cnt = 0; q = p->next; while(q && cnt < k-2){ q = q->next; cnt++; } if(q == NULL){ //不够K个数 return head; } cnt = 0; while( cnt < k-1){ q = p->next; p->next = q->next; if(isHead){ q->next = head; head = q; }else{ q->next = prev->next; prev->next = q; } cnt++; } isHead = false; prev = p; p = p->next; } return head; } };