【PAT甲级 链表处理】1097 Deduplication on a Linked List (25 分)

更改链表结构

# include<bits/stdc++.h>
using namespace std;

struct Node{
    
    
    int data;
    int next;
}node[100010];

int start, N;
bool visit[100010];

int main() {
    
    
    cin >> start >> N;
    for(int i = 0;i < N;++i){
    
    
        int now;
        cin >> now;
        cin >> node[now].data >> node[now].next;
        visit[abs(node[now].data)] = false;
    }

	// 原地更改链表结构
    int rmv_start = -1, rebuild_tail = start, rmv_tail = -1;
    int now;
    for(now = start;now != -1;now = node[now].next){
    
    
        if(visit[abs(node[now].data)] == true){
    
     // 是重复的，所以要remove掉
            if(rmv_start == -1){
    
     
                rmv_start = now; // 这是remove链的头
            } else {
    
     
                node[rmv_tail].next = now; // 将重复结点连上remove链
            }
            rmv_tail = now; // 更新remove链的尾结点
        } else {
    
     // 是第一次出现的元素, 不属于remove链
            visit[abs(node[now].data)] = true;
            if (now != start) 
				node[rebuild_tail].next = now;
            rebuild_tail = now; // 确认now结点不属于remove链
        }
    }
    node[rebuild_tail].next = node[rmv_tail].next = -1;



    for(int p = start;p != -1;p = node[p].next){
    
    
        printf("%05d %d ", p, node[p].data);
        if(node[p].next == -1) 
            cout << -1 << endl;
        else 
            printf("%05d\n", node[p].next);
    }
    for(int p = rmv_start;p != -1;p = node[p].next){
    
    
        printf("%05d %d ", p, node[p].data);
        if(node[p].next == -1) 
            cout << -1 << endl;
        else 
            printf("%05d\n", node[p].next);
    }

    return 0;
}

---

不更改链表结构，将链表按顺序存入线性表

# include<bits/stdc++.h>
using namespace std;

struct Node{
    
    
    int data;
    int next;
}node[100010];

int start, N;
bool visit[100010] = {
    
    0};

int main() {
    
    
	// 输入，建立静态链表
    cin >> start >> N;
    for(int i = 0;i < N;++i){
    
    
        int now;
        cin >> now;
        cin >> node[now].data >> node[now].next;
    }

    // 将链表变成顺序表，可以简化非常多！！
    vector<int> V, V_dep;
    for(int now = start;now != -1;now = node[now].next){
    
    
        if(visit[abs(node[now].data)] == false){
    
    
            V.push_back(now);
            visit[abs(node[now].data)] = true;
        } else {
    
    
            V_dep.push_back(now);
        }
    }

	// 输出时真正的next是顺序表的下一个元素
    for(int i = 0;i < V.size();++i){
    
    
        int now = V[i];
        if(i != V.size() - 1){
    
    
            int next = V[i+1];
            printf("%05d %d %05d\n", now, node[now].data, next);
        }
        else
            printf("%05d %d %d\n", now, node[now].data, -1);
    }
    for(int i = 0;i < V_dep.size();++i){
    
    
        int now = V_dep[i];
        if(i != V_dep.size() - 1){
    
    
            int next = V_dep[i+1];
            printf("%05d %d %05d\n", now, node[now].data, next);
        }
        else
            printf("%05d %d %d\n", now, node[now].data, -1);
    }

    return 0;
}