CC10. 牛牛的双链表求和
描述
输入描述
输出描述
把数组 a 和数组 b 转换成链表，然后把链表 a 中的值加到链表 b 中，然后输出加和后的链表。示例1
输入:
5 5 4 2 1 3 2 4 5 8 9
输出:
7 8 7 9 12
C 解法, 执行用时: 2ms, 内存消耗: 296KB, 提交时间: 2022-03-14
#include <stdio.h>
#include <stdlib.h>
typedef struct list{
int data;
struct list *next;
}list,*linklist;
//链表内容打印
void ptink(list *head)
{
struct list *point=head;
while(point!=NULL)
{
printf("%d ",point->data);
point=point->next;//while里面的判断其实是在判断point->next是否为NULL
}
putchar('\n');
}
//头插法
list *headInsert(list *head1,int *arr1,int n)
{
int i;
list *headList = head1;
list *arrList = NULL;
for(i=n-1;i>=0;i--)
{
arrList = (list *)malloc(sizeof(list));
arrList->data = arr1[i];
arrList->next = headList;
headList = arrList;
}
head1 = headList;
return head1;
}
//尾插法
list *trailInsert(list *head2,int *arr2,int n)
{
int i;
list *headList = head2;
list *arrList = NULL;
for(i=0;i<n;i++)
{
arrList = (list *)malloc(sizeof(list));
arrList->data = arr2[i];
arrList->next = NULL; //将headList放在arrList->next,即arrList的数据是放在headList的前面的
//headList=head;
if(headList == NULL)
{
headList = arrList; //
head2 = headList;
}
else
{
while(headList->next != NULL)
{
headList = headList->next ;
}
headList->next = arrList;
}
}
return head2;
}
//求和链表
list *Sum(list *head,list *head1,list *head2)
{
list *List = NULL; //每个节点存数的
list *list0 = head; //临时头链表
list *list1 = head1;
list *list2 = head2;
while(list1 != NULL)
{
List = (list *)malloc(sizeof(list));
List->data = (list1->data)+ (list2->data) ;
List->next =NULL;
//尾插法
if(list0 == NULL)
{
list0 = List; //
head = list0;
}
else
{
while(list0->next !=NULL)
{
list0 = list0->next; //链表往后位移
}
list0->next = List;
}
list1 = list1->next;
list2 = list2->next;
}
return head;
}
int main()
{
int i,n;
scanf("%d",&n);
int arr1[n];
int arr2[n];
for(i=0;i<n;i++)
{
scanf("%d",&arr1[i]);
}
for(i=0;i<n;i++)
{
scanf("%d",&arr2[i]);
}
list *head = NULL;
list *head1 = NULL;
list *head2 = NULL;
head1 = headInsert(head1,arr1,n);
//ptink(head1);
head2 = trailInsert(head2,arr2,n);
//ptink(head2);
head = Sum( head, head1, head2);
ptink(head);
return 0;
}
C 解法, 执行用时: 2ms, 内存消耗: 304KB, 提交时间: 2022-04-07
#include <stdio.h>
#include <stdlib.h>
typedef struct list{
int data;
struct list *next;
}list,*linklist;
//链表内容打印
void ptink(list *head)
{
struct list *point=head;
while(point!=NULL)
{
printf("%d ",point->data);
point=point->next;//while里面的判断其实是在判断point->next是否为NULL
}
putchar('\n');
}
//头插法
list *headInsert(list *head1,int *arr1,int n)
{
int i;
list *headList = head1;
list *arrList = NULL;
for(i=n-1;i>=0;i--)
{
arrList = (list *)malloc(sizeof(list));
arrList->data = arr1[i];
arrList->next = headList;
headList = arrList;
}
head1 = headList;
return head1;
}
//尾插法
list *trailInsert(list *head2,int *arr2,int n)
{
int i;
list *headList = head2;
list *arrList = NULL;
for(i=0;i<n;i++)
{
arrList = (list *)malloc(sizeof(list));
arrList->data = arr2[i];
arrList->next = NULL; //将headList放在arrList->next,即arrList的数据是放在headList的前面的
//headList=head;
if(headList == NULL)
{
headList = arrList; //
head2 = headList;
}
else
{
while(headList->next != NULL)
{
headList = headList->next ;
}
headList->next = arrList;
}
}
return head2;
}
//求和链表
list *Sum(list *head,list *head1,list *head2)
{
list *List = NULL; //每个节点存数的
list *list0 = head; //临时头链表
list *list1 = head1;
list *list2 = head2;
while(list1 != NULL)
{
List = (list *)malloc(sizeof(list));
List->data = (list1->data)+ (list2->data) ;
List->next =NULL;
//尾插法
if(list0 == NULL)
{
list0 = List; //
head = list0;
}
else
{
while(list0->next !=NULL)
{
list0 = list0->next; //链表往后位移
}
list0->next = List;
}
list1 = list1->next;
list2 = list2->next;
}
return head;
}
int main()
{
int i,n;
scanf("%d",&n);
int arr1[n];
int arr2[n];
for(i=0;i<n;i++)
{
scanf("%d",&arr1[i]);
}
for(i=0;i<n;i++)
{
scanf("%d",&arr2[i]);
}
list *head = NULL;
list *head1 = NULL;
list *head2 = NULL;
head1 = headInsert(head1,arr1,n);
//ptink(head1);
head2 = trailInsert(head2,arr2,n);
//ptink(head2);
head = Sum( head, head1, head2);
ptink(head);
return 0;
}
// #include <stdio.h>
// #include <stdlib.h>
// typedef struct link{
// int data;
// struct link *next;
// }link, *linklist;
// void SumList(link *la, link *lb)
// {
// link *lc = (link *)malloc(sizeof(link));
// link *pc = lc;
// while(la->next != NULL)
// {
// link *temp_list = (link *)malloc(sizeof(link));
// temp_list->data = la->data + lb->data;
// temp_list->next = NULL;
// pc->next = temp_list;
// pc = pc->next;
// la = la->next;
// lb = lb->next;
// }
// pc = lc;
// while(pc->next != NULL)
// {
// printf("%d ", pc->data);
// pc = pc->next;
// }
// }
// int main()
// {
// int i, n;
// int temp;
// scanf("%d", &n);
// link *heada = (link *)malloc(sizeof(link));
// link *pa = heada;
// link *headb = (link *)malloc(sizeof(link));
// link *pb = heada;
// for(i = 0; i<n; i++)
// {
// scanf("%d", &temp);
// link *alist = (link *)malloc(sizeof(link));
// alist->data = temp;
// alist->next = NULL;
// pa->next = alist;
// pa = pa->next;
// }
// for(i=0; i<n; i++)
// {
// scanf("%d", &temp);
// link *blist = (link *)malloc(sizeof(link));
// blist->data = temp;
// blist->next = NULL;
// pb->next = blist;
// pb = pb->next;
// }
// SumList(heada, headb);
// return 0;
// }
C 解法, 执行用时: 2ms, 内存消耗: 312KB, 提交时间: 2022-07-29
#include <stdio.h>
#include <stdlib.h>
typedef struct Node
{
int num;
struct Node *next;
}LinkNode;
LinkNode *creat_link(int n){
int i;
LinkNode *head = (LinkNode *)malloc(sizeof(LinkNode));
LinkNode *tail;
head->next = NULL;
scanf("%d", &head->num);
tail = head;
for (i = 0; i < n-1; ++i)
{
LinkNode *p = (LinkNode *)malloc(sizeof(LinkNode));
scanf("%d", &p->num);
p->next = NULL;
tail->next = p;
tail = tail->next;
}
return head;
}
void add_node(int n, const LinkNode *head_1, LinkNode *head_2){
int i;
LinkNode *tail_1 = head_1;
LinkNode *tail_2 = head_2;
for(i = 0; i < n; ++i){
tail_2->num = tail_2->num + tail_1->num;
tail_1 = tail_1->next;
tail_2 = tail_2->next;
}
}
int main(int argc, char const *argv[])
{
int n,i;
scanf("%d", &n);
LinkNode *head_1 = creat_link(n);
LinkNode *tail;
LinkNode *head_2 = creat_link(n);
add_node(n, head_1, head_2);
tail = head_2;
while(tail != NULL){
printf("%d ", tail->num);
tail = tail->next;
}
printf("\n");
return 0;
}
C 解法, 执行用时: 2ms, 内存消耗: 320KB, 提交时间: 2022-07-30
#include<stdio.h>
typedef struct Node
{
int value;
struct Node *link;
}node;
node*Init_list(int n)
{
node*header=(node*)malloc(sizeof(node));
header->value=0;
header->link=NULL;
node*pRear=header;
int count=0;
int v;
while(count<n)
{
scanf("%d",&v);
node*code=(node*)malloc(sizeof(node));
code->value=v;
code->link=NULL;
pRear->link=code;
pRear=code;
count++;
}
return header;
}
int main()
{
int n;
scanf("%d",&n);
node*p=Init_list(n)->link;
node*q=Init_list(n)->link;
while(p!=NULL)
{
p->value+=q->value;
printf("%d ",p->value);
p=p->link;
q=q->link;
}
return 0;
}
C 解法, 执行用时: 2ms, 内存消耗: 320KB, 提交时间: 2022-03-08
#include <stdio.h>
#include <stdlib.h>
typedef struct list{
int data;
struct list *next;
}list,*linklist;
//链表内容打印
void ptink(list *head)
{
struct list *point=head;
while(point!=NULL)
{
printf("%d ",point->data);
point=point->next;//while里面的判断其实是在判断point->next是否为NULL
}
putchar('\n');
}
#if 1
//头插法
list *headInsert(list *head1,int *arr1,int n)
{
int i;
list *headList = head1;
list *arrList = NULL;
for(i=n-1;i>=0;i--)
{
arrList = (list *)malloc(sizeof(list));
arrList->data = arr1[i];
arrList->next = headList;
headList = arrList;
}
head1 = headList;
return head1;
}
//尾插法
list *trailInsert(list *head2,int *arr2,int n)
{
int i;
list *headList = head2;
list *arrList = NULL;
for(i=0;i<n;i++)
{
arrList = (list *)malloc(sizeof(list));
arrList->data = arr2[i];
arrList->next = NULL; //将headList放在arrList->next,即arrList的数据是放在headList的前面的
//headList=head;
if(headList == NULL)
{
headList = arrList; //
head2 = headList;
}
else
{
while(headList->next != NULL)
{
headList = headList->next ;
}
headList->next = arrList;
}
}
return head2;
}
//求和链表
list *Sum(list *head,list *head1,list *head2)
{
list *List = NULL; //每个节点存数的
list *list0 = head; //临时头链表
list *list1 = head1;
list *list2 = head2;
while(list1 != NULL)
{
List = (list *)malloc(sizeof(list));
List->data = (list1->data)+ (list2->data) ;
List->next =NULL;
//尾插法
if(list0 == NULL)
{
list0 = List; //
head = list0;
}
else
{
while(list0->next !=NULL)
{
list0 = list0->next; //链表往后位移
}
list0->next = List;
}
list1 = list1->next;
list2 = list2->next;
}
return head;
}
int main()
{
int i,n;
scanf("%d",&n);
int arr1[n];
int arr2[n];
for(i=0;i<n;i++)
{
scanf("%d",&arr1[i]);
}
for(i=0;i<n;i++)
{
scanf("%d",&arr2[i]);
}
list *head = NULL;
list *head1 = NULL;
list *head2 = NULL;
head1 = headInsert(head1,arr1,n);
//ptink(head1);
head2 = trailInsert(head2,arr2,n);
//ptink(head2);
head = Sum( head, head1, head2);
ptink(head);
return 0;
}
#endif