Given two singly linked lists L1=a1→a2→⋯→an−1→an and L2=b1→b2→⋯→bm−1→bm. If n≥2m, you are supposed to reverse and merge the shorter one into the longer one to obtain a list like a1→a2→bm→a3→a4→bm−1⋯. For example, given one list being 6→7 and the other one 1→2→3→4→5, you must output 1→2→7→3→4→6→5.
Input Specification:
Each input file contains one test case. For each case, the first line contains the two addresses of the first nodes of L1 and L2, plus a positive N (≤105) which is the total number of nodes given. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.
Then N lines follow, each describes a node in the format:
Address Data Next
where Address is the position of the node, Data is a positive integer no more than 105, and Next is the position of the next node. It is guaranteed that no list is empty, and the longer list is at least twice as long as the shorter one.
Output Specification:
For each case, output in order the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.
Sample Input:
00100 01000 7
02233 2 34891
00100 6 00001
34891 3 10086
01000 1 02233
00033 5 -1
10086 4 00033
00001 7 -1
Sample Output:
01000 1 02233
02233 2 00001
00001 7 34891
34891 3 10086
10086 4 00100
00100 6 00033
00033 5 -1题目大意：给定两个单链表L1 = a1 → a2 → … → an-1 → an，和L2 = b1 → b2 → … → bm-1 → bm，将较短的那个链表逆序，然后将之并入比较长的那个链表，得到一个形如a1 → a2 → bm → a3 → a4 → bm-1 … 的结果，例如给定两个链表分别为6→7和1→2→3→4→5，你应该输出1→2→7→3→4→6→5。
分析：由于题目没有说哪个链表更长，因此首先要先确定短的链表是哪个，再把短的链表逆序。之后长的链表每前进2步，短的链表前进1步。注意有可能n=2*m的边界条件。
#include<algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <ctime>
#include <cmath>
#include <map>
#include <set>
#define INF 0xffffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
using namespace std;
typedef struct node
{
int val,next,id;
}node;
int main(void)
{
#ifdef test
freopen("in.txt","r",stdin);
//freopen("in.txt","w",stdout);
clock_t start=clock();
#endif //test
int r1,r2,n;scanf("%d%d%d",&r1,&r2,&n);
node num[100000];
for(int i=0;i<100000;++i)
num[i].val=0,num[i].next=-1,num[i].id=i;
for(int i=0;i<n;++i)
{
int a,b,c;scanf("%d%d%d",&a,&b,&c);
num[a].val=b,num[a].next=c;
}
int t1=0,t2=0,pos1=r1,pos2=r2;
while(pos1!=-1)
t1++,pos1=num[pos1].next;
while(pos2!=-1)
t2++,pos2=num[pos2].next;
if(t1<t2)swap(r1,r2);
pos1=r1,pos2=r2;
int temp=-1,next=-1;
while(pos2!=-1)
{
if(num[pos2].next==-1)r2=pos2;
next=num[pos2].next,num[pos2].next=temp,temp=pos2,pos2=next;
}
// pos2=r2;
// while(pos2!=-1)
// {
// printf("%05d %d %05d\n",num[pos2].id,num[pos2].val,num[pos2].next);
// pos2=num[pos2].next;
// }printf("\n");
pos2=r2;
int t=0;
while(pos1!=-1)
{
if(t<2||pos2==-1)
{
if(pos1==r1)printf("%05d %d",num[pos1].id,num[pos1].val);
else printf(" %05d\n%05d %d",num[pos1].id,num[pos1].id,num[pos1].val);
t++;pos1=num[pos1].next;
}
else
{
printf(" %05d\n%05d %d",num[pos2].id,num[pos2].id,num[pos2].val);pos2=num[pos2].next;t=0;
}
}
if(pos2!=-1)printf(" %05d\n%05d %d",num[pos2].id,num[pos2].id,num[pos2].val);pos2=num[pos2].next;t=0;
printf(" -1\n");
#ifdef test
clockid_t end=clock();
double endtime=(double)(end-start)/CLOCKS_PER_SEC;
printf("\n\n\n\n\n");
cout<<"Total time:"<<endtime<<"s"<<endl; //s为单位
cout<<"Total time:"<<endtime*1000<<"ms"<<endl; //ms为单位
#endif //test
return 0;
}
