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
分析
屬於甲級常考的靜態鏈表題目。如果想了解靜態鏈表題的思路可以先去看晴神寶典。
第一種寫法就是晴神寶典的寫法，用排序，只用一個vector，並且會修改到新鏈表的next項。
第二種寫法就是參考1133柳神的寫法，不同排序，用多個vector，不用修改next項。
滿分代碼（方法一）
#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <cmath>
using namespace std;
int k, n,m,a,b,c,d,e,ar=1,br=1;
int ad[1000005];
struct node{
int addr,data,next,rank;
bool f=false;
};
vector<node> v;
bool cmp(node &a,node &b){
if(a.f!=b.f)return a.f>b.f;
else return a.rank<b.rank;
}
int main() {
scanf("%d %d %d", &a,&b,&n);
v.resize(n);
for(int i=0;i<n;i++){
scanf("%d %d %d",&c,&d,&e);
ad[c]=i;
v[i].addr=c;
v[i].data=d;
v[i].next=e;
}
int as=a,bs=b;
while(as!=-1){
v[ad[as]].rank=ar;
v[ad[as]].f=true;
as=v[ad[as]].next;
ar++;
}
br=ar;
while(bs!=-1){
v[ad[bs]].rank=br;
v[ad[bs]].f=true;
bs=v[ad[bs]].next;
br++;
}
ar--;br--;
if(ar<br-ar){
as=b;
bs=a;
ar=1;
while(as!=-1){
v[ad[as]].rank=ar;
v[ad[as]].f=true;
as=v[ad[as]].next;
ar++;
}
br=ar;
while(bs!=-1){
v[ad[bs]].rank=br;
v[ad[bs]].f=true;
bs=v[ad[bs]].next;
br++;
}
ar--;br--;
}
sort(v.begin(),v.end(),cmp);
for(int i=0;i<br;i++){
if(i>1 && i<ar){
v[i].rank+=i/2;
}else if(i>=ar){
v[i].rank=(br-i)*3;
}
}
sort(v.begin(),v.end(),cmp);
int bl=br-ar;
for(int i=0;i<br;i++){
if(bl>0 && i>0 && (i+1)%3==0){
int t=v[i-1].next;
v[i-1].next=v[i].addr;
v[i].next=t;
bl--;
}
}
for(int i=0;i<br;i++){
if(i==br-1) printf("%05d %d -1",v[i].addr,v[i].data);
else printf("%05d %d %05d\n",v[i].addr,v[i].data,v[i].next);
}
return 0;
}
滿分代碼（方法二）
#include<iostream>
#include<vector>
using namespace std;
struct node{
int id,data,next;
};
node a[1000005];
int main(){
int ba,bb,n,s,d,e;
scanf("%d %d %d\n",&ba,&bb,&n);
vector<node> va,vb,ans;
for(int i=0;i<n;i++){
scanf("%d %d %d\n",&s,&d,&e);
a[s]={s,d,e};
}
for(;ba!=-1;ba=a[ba].next){
va.push_back(a[ba]);
}
for(;bb!=-1;bb=a[bb].next){
vb.push_back(a[bb]);
}
if(va.size()>vb.size()){
int j=(int)vb.size()-1;
for(int i=0;i<va.size();i++){
if(i>0 && i%2==0 && j>=0){
ans.push_back(vb[j]);
j--;
}
ans.push_back(va[i]);
if(i==va.size()-1 && j==0){
ans.push_back(vb[j]);
}
}
}else {
int j=(int)va.size()-1;
for(int i=0;i<vb.size();i++){
if(i>0 && i%2==0 && j>=0){
ans.push_back(va[j]);
j--;
}
ans.push_back(vb[i]);
if(i==vb.size()-1 && j==0){
ans.push_back(va[j]);
}
}
}
for(int i=0;i<ans.size();i++){
if(i!=ans.size()-1)printf("%05d %d %05d\n",ans[i].id,ans[i].data,ans[i+1].id);
else printf("%05d %d -1",ans[i].id,ans[i].data);
}
return 0;
}
更多測試數據集
//輸入1
00100 01000 6
02233 2 34891
00100 6 00001
34891 3 10086
01000 1 02233
10086 4 -1
00001 7 -1
//輸出1
01000 1 02233
02233 2 00001
00001 7 34891
34891 3 10086
10086 4 00100
00100 6 -1
//輸入2
00100 01000 8
02233 2 34891
00100 7 00001
34891 3 10086
01000 1 02233
00033 5 02342
10086 4 00033
00001 8 -1
02342 6 -1
//輸出2
01000 1 02233
02233 2 00001
00001 8 34891
34891 3 10086
10086 4 00100
00100 7 00033
00033 5 02342
02342 6 -1
//輸入3
01000 00033 3
02233 2 -1
01000 1 02233
00033 3 -1
//輸出3
01000 1 02233
02233 2 00033
00033 3 -1
//輸入4
00033 01000 3
02233 2 -1
01000 1 02233
00033 3 -1
//輸出4
01000 1 02233
02233 2 00033
00033 3 -1
//輸入6:遊離節點不能輸出
02233 01000 6
02233 2 00033
01000 1 -1
00033 3 23421
23421 4 -1
27834 9 78932
78932 8 -1
//輸出6
02233 2 00033
00033 3 01000
01000 1 23421
23421 4 -1