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;
}