PAT甲级专题|链表
PAT链表专题
关于PAT甲级的链表问题，主要内容 就是”建立链表“
所以第一步学会模拟链表，pat又不卡时间，这里用vector + 结构体，更简洁
模拟链表的普遍代码
const int maxn = 1e6+10;
struct node{
int address;
int next;
char key;
}nod[maxn];
int head1,n;
vector<node> list1;
cin>>head1>>n;
for(int i=1;i<=n;i++) {
int address,next;
char key;
cin>>address>>key>>next;
nod[address].address = address;//重点1
nod[address].key = key;
nod[address].next = next;
}
for(head1; head1 != -1; head1 = nod[head1].next){ //重点2
list1.push_back(nod[head1]);
}
学会模拟链表之后，PAT甲级的链表题就都能做了，万变不离其宗，
基本就是，建立链表、按照题意操作链表（链表合并、去重、反转...）、注意判断链表边界（链表有效长度）、输入输出( %05d 来输出地址)
2019秋季PAT 7-2
#include<bits/stdc++.h>
using namespace std;
const int maxn = 100001;
int ad1,ad2,n;
int head1,head2 = 0;
int len1 = 0,len2 = 0;
struct node{
int add,next,key;
}nod[maxn];
vector<node> list1;
vector<node> list2;
vector<node> res;
int main(){
cin>>head1>>head2>>n;
for(int i=1;i<=n;i++){
int add,next,key;
cin>>add>>key>>next;
nod[add].add = add;
nod[add].next = next;
nod[add].key = key;
}
for (int p = head1; p != -1; p = nod[p].next)
list1.push_back(nod[p]);
for(int p = head2;p!=-1;p = nod[p].next)
list2.push_back(nod[p]);
len1 = list1.size()-1;
len2 = list2.size()-1;
if(len1 >= 2*len2){
int p = 0,q = len2;
while(q >= 0){
res.push_back(list1[p++]);
res.push_back(list1[p++]);
res.push_back(list2[len2--]);
}
while(p <= len1){
res.push_back(list1[p++]);
}
for(int i=0;i<res.size();i++){
if(i != res.size() - 1){
res[i].next = res[i+1].add;
}
}
}else{
int p = len1,q = 0;
while(p >= 0){
res.push_back(list2[q++]);
res.push_back(list2[q++]);
res.push_back(list1[p--]);
}
while(q<=len2){
res.push_back(list2[q++]);
}
for(int i=0;i<res.size();i++){
if(i != res.size() - 1){
res[i].next = res[i+1].add;
}
}
}
for(int i=0;i<res.size();i++){
if(i == res.size()-1) printf("%05d %d -1\n",res[i].add,res[i].key);
else printf("%05d %d %05d\n",res[i].add,res[i].key,res[i].next);
}
return 0;
}
PTA1032
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+10;
struct node{
int address;
int next;
char key;
}nod[maxn];
int head1,head2,n;
bool use[maxn];
vector<node> list1,list2;
int main(){
cin>>head1>>head2>>n;
for(int i=1;i<=n;i++) {
int address,next;
char key;
cin>>address>>key>>next;
nod[address].address = address;
nod[address].key = key;
nod[address].next = next;
}
for(head1; head1 != -1; head1 = nod[head1].next){
list1.push_back(nod[head1]);
use[head1] = true;
}
for(head2; head2 != -1; head2 = nod[head2].next){
if(use[head2]){
printf("%05d\n",head2);
return 0;
}
}
cout<<"-1"<<endl;
return 0;
}
PTA1052
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+100;
struct node{
int address;
int key;
int next;
bool flag;
}nod[100010];
vector<node> list1;
int n;
int head;
bool cmp(node a,node b){
return a.key < b.key;
}
int main(){
cin>>n>>head;
for(int i=1;i<=n;i++){
int address,key,next;
cin>>address>>key>>next;
nod[address].address = address;
nod[address].key = key;
nod[address].next = next;
nod[address].flag = true;
}
int m = 0;
for(;head!=-1;head = nod[head].next){
nod[head].flag = true;
list1.push_back(nod[head]);
m++;
}
if(m == 0){ //判边界 否则段错误
printf("0 -1\n");
return 0;
}
sort(list1.begin(),list1.end(),cmp);
for(int i=0;i<m-1;i++){
list1[i].next = list1[i+1].address;
}
cout<<m<<" ";
printf("%05d\n",list1[0].address);
for(int i=0;i<m-1;i++){
printf("%05d %d %05d\n",list1[i].address,list1[i].key,list1[i].next);
}
printf("%05d %d -1\n",list1[m-1].address,list1[m-1].key);
return 0;
}
PTA1074
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
/*
题目意思是 k个元素间换一下
*/
struct node{
int address;
int key;
int next;
}nod[maxn];
vector<node>list1,list2;
int head,n,k;
void print(){
for(int i=0;i<n-1;i++){
printf("%05d %d %05d\n",list2[i].address,list2[i].key,list2[i].next);
}
printf("%05d %d -1\n",list2[n-1].address,list2[n-1].key);
}
int main(){
cin>>head>>n>>k;
for(int i=0;i<n;i++){
int add,key,next;
cin>>add>>key>>next;
nod[add].address = add;
nod[add].key = key;
nod[add].next = next;
}
for(;head!=-1;head=nod[head].next)
list1.push_back(nod[head]);
n = list1.size(); //原来的n不一定是新链表的长度 可能输入数据有无效结点
int pos = 0;
while(pos + k - 1 < n){ //这里注意 下标 每pos~pos+k-1为一组
for(int i = pos+k-1;i>=pos;i--){ //倒序存进list2
list2.push_back(list1[i]);
}
pos+=k;
}
if(pos < n){ //剩下的没有凑够k个就 直接存入链表list2
for(int i=pos;i<n;i++){//正序存进list2
list2.push_back(list1[i]);
}
}
for(int i=0;i<n-1;i++){ //更新next地址
list2[i].next = list2[i+1].address;
}
list2[n-1].next = -1;
print();
return 0;
}
/*
00100 4 2
00000 4 -1
00100 1 12309
33218 3 00000
12309 2 33218
*/
PTA1097
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
int head,n;
struct node{
int address;
int key;
int next;
}nod[maxn];
bool vis[maxn];
vector<node> list1,list2;
int main(){
cin>>head>>n;
for(int i = 0;i < n;i++){
int add,key,next;
cin>>add>>key>>next;
nod[add].address = add;
nod[add].key = key;
nod[add].next = next;
}
for(;head!=-1;head = nod[head].next){
if(!vis[abs(nod[head].key)]){
vis[abs(nod[head].key)] = true;
list1.push_back(nod[head]);
}else{
list2.push_back(nod[head]);
}
}
n = list1.size();
if(n != 0){
for(int i=0;i<n-1;i++){
printf("%05d %d %05d\n",list1[i].address,list1[i].key,list1[i+1].address);
}
printf("%05d %d -1\n",list1[n-1].address,list1[n-1].key);
}
int m = list2.size();
if(m != 0){
for(int i=0;i<m-1;i++){
printf("%05d %d %05d\n",list2[i].address,list2[i].key,list2[i+1].address);
}
printf("%05d %d -1\n",list2[m-1].address,list2[m-1].key);
}
return 0;
}
