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;
} 

标签：head,PAT,int,next,链表,nod,甲级,key,address
来源： https://www.cnblogs.com/fisherss/p/11516227.html