标签：png name extra 较差 content 内存 span space solution
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
Example:
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
Note:
改进声明：
本方法速度和内存好像都比较差，虽然思路简洁，但是可能构造技巧上还是有进步空间
main idea:
One word to sum up the idea,victory! haaa~just a joke
Group
In this reverse problem,group is important,devide into ordered &inordered.three pointers we need to correctly understanding the meaning of these variables.cur->ordered part end ,nxt->inordered start,pre->before these parts.
It looks like this composition:
pre->order part->inorder part->left_out
Now it‘s the code
#include <iostream> #include<vector> #include<iostream> #include<string> #include<stdio.h> #include<string.h> #include<iomanip> #include<vector> #include<list> #include<queue> #include<algorithm> #include<stack> #include<map> using namespace std; struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: ListNode* reverseKGroup(ListNode* head, int k) { //this situation can include head->next //special case dispose if(!head||k==1) return head; //dummy node ListNode dummy(0); dummy.next=head; //len calc int len=1; while(head=head->next) len++; //cout<<len<<endl; ListNode *pre=&dummy; for(int i=0;i+k<=len;i+=k) { ListNode *cur=pre->next; ListNode *nxt=cur->next; for(int j=1;j<k;j++) { cur->next=nxt->next; nxt->next=pre->next; pre->next=nxt; nxt=cur->next; } pre=cur; } return dummy.next; } }; int main() { int k; cin>>k; ListNode a(1); ListNode b(2); ListNode c(3); ListNode d(4); ListNode e(5); Solution s; a.next=&b; b.next=&c; c.next=&d; d.next=&e; ListNode *head=&a; // while(head) // { // cout<<head->val<<endl; // head=head->next; // } ListNode* res=NULL; res=s.reverseKGroup(head, k); while(res) { cout<<res->val<<endl; res=res->next; } return 0; }
LeetCode开心刷题十四天——25Reverse Nodes in k-Group
标签：png name extra 较差 content 内存 span space solution
原文地址：https://www.cnblogs.com/Marigolci/p/11145448.html
