作者:
晚于: 2020-07-08 12:00:00后提交分数乘系数50%
截止日期: 2020-07-15 12:00:00
问题描述 :
给定一个单链表 L:L0→L1→…→Ln-1→Ln ,
将其重新排列后变为: L0→Ln→L1→Ln-1→L2→Ln-2→…
你不能只是单纯的改变节点内部的值,而是需要实际的进行节点交换。
示例 1:
给定链表 1->2->3->4, 重新排列为 1->4->2->3.
示例 2:
给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.
可使用以下代码,完成其中的reorderList函数,其中形参head指向无头结点单链表。
#include<iostream>
using namespace std;
struct ListNode
{
int val;
ListNode *next;
ListNode() : val(0), next(NULL) {}
ListNode(int x) : val(x), next(NULL) {}
ListNode(int x, ListNode *next) : val(x), next(next) {}
};
class Solution
{
public:
void reorderList(ListNode* head)
{
//填充本函数完成功能
}
};
ListNode *createByTail()
{
ListNode *head;
ListNode *p1,*p2;
int n=0,num;
int len;
cin>>len;
head=NULL;
while(n<len && cin>>num)
{
p1=new ListNode(num);
n=n+1;
if(n==1)
head=p1;
else
p2->next=p1;
p2=p1;
}
return head;
}
void displayLink(ListNode *head)
{
ListNode *p;
p=head;
cout<<"head-->";
while(p!= NULL)
{
cout<<p->val<<"-->";
p=p->next;
}
cout<<"tail ";
}
int main()
{
ListNode* head = createByTail();
Solution().reorderList(head);
displayLink(head);
return 0;
}
#include<iostream> using namespace std; struct ListNode { int val; ListNode *next; ListNode() : val(0), next(NULL) {} ListNode(int x) : val(x), next(NULL) {} ListNode(int x, ListNode *next) : val(x), next(next) {} }; class Solution//将链表分两半,将第二个逆序,在依次连接 { public: void reorderList(ListNode* head) { if (!head || !head->next || !head->next->next) return; //找中点,链表分成两个 ListNode *s = head; ListNode *f = head; while (f->next && f->next->next) { s = s->next; f = f->next->next; } ListNode *newHead = s->next; s->next = NULL; //第二个链表倒置 newHead = reverseList(newHead); //链表节点依次连接 while (newHead) { ListNode* temp = newHead->next; newHead->next = head->next; head->next = newHead; head = newHead->next; newHead = temp; } } private: ListNode* reverseList(ListNode* head) { if (!head) return NULL; ListNode* tail = head; head = head->next; tail->next = NULL; while (head) { ListNode* temp = head->next; head->next = tail; tail = head; head = temp; } return tail; } }; ListNode *createByTail() { ListNode *head; ListNode *p1,*p2; int n=0,num; int len; cin>>len; head=NULL; while(n<len && cin>>num) { p1=new ListNode(num); n=n+1; if(n==1) head=p1; else p2->next=p1; p2=p1; } return head; } void displayLink(ListNode *head) { ListNode *p; p=head; cout<<"head-->"; while(p!= NULL) { cout<<p->val<<"-->"; p=p->next; } cout<<"tail "; } int main() { ListNode* head = createByTail(); Solution().reorderList(head); displayLink(head); return 0; }
思想:将链表分两半,将第二个逆序,在依次连接