首页 > 代码库 > Leetcode--Reorder List
Leetcode--Reorder List
Problem Description:
Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
You must do this in-place without altering the nodes‘ values.
For example,
Given {1,2,3,4}
, reorder it to {1,4,2,3}
.
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: void reorderList(ListNode *head) { ListNode *p=head; int length,i=1,tag=0; vector<ListNode* > array; while(p) { array.push_back(p); p=p->next; } p=head; if(array.size()<3) return; length=array.size()-1; while(tag<array.size()/2) { p->next=array[length--]; p=p->next; p->next=array[i++]; p=p->next; tag++; } p->next=NULL; } };
第二种方法就是找到链表的中间节点,将链表分为两半,然后将后半部分转置,最后将前后两部分进行合并即可,具体代码如下:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: void reorderList(ListNode *head) { if (!head){return;} if (head->next==NULL){return;} ListNode *p=head; ListNode *q=head; //find the midddle pointer while (q->next && q->next->next){ p=p->next; q=q->next->next; } //now p is middle pointer //reverse p->next to end q = p->next; while (q->next){ ListNode* tmp = p->next; p->next = q->next; q->next = q->next->next; p->next->next = tmp; } //reorder q = head; while (p!=q && p->next){ ListNode* tmp = q->next; q->next = p->next; p->next = p->next->next; q->next->next = tmp; q=q->next->next; } return; } };
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。