25. Reverse Nodes in k-Group
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LeetCode链接
描述:给一个链表,对间隔为k个元素翻转
要求:不能改变链表的值,
空间复杂度为常量
例子:
指定链表:1->2->3->4->5
k=2,输出:2->1->4->3->5
k=3,输出:3->2->1->4->5
class Solution {public: ListNode* reverseKGroup(ListNode* head, int k) { //防止头指针丢失 ListNode* guard=new ListNode(0); guard->next=head; head=guard; for(;;){ int res_node=0; //要交换的第k个元素的下一个 ListNode* last=head->next; //统计链表还剩多少个 for (int j=0;j<k;j++){ if(last==NULL) break; res_node+=1; last=last->next; } if(res_node<k) break; //要交换的第一个元素 ListNode*first=head->next; for(int j=0;j<k;j++){ int index=k-j-1; ListNode*aim=first; while(index--){ aim=aim->next; } //把第K个节点放到前面去 head->next=aim; head=aim; } head->next=last; } return guard->next; }};
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