天梯赛 L2-022 重排链表

使用两个map记录每个地址对应的值和每个地址对应的下一个地址，然后模拟即可

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long
typedef long long ll;
const int N = 3e5 + 10;
const int mod = 998244353;
void solve() {
	string st;
	int n;
	cin>>st>>n;
	map<string,int> value;	//地址对应的值
	map<string ,string> ne;	//该地址的下一个地址
	for(int i = 1 ; i <= n ; i++){
		string address;
		int data;
		string next;
		cin>>address>>data>>next;
		value[address] = data;
		ne[address] = next;
	}
	string now = st;
	vector<pair<string,int>> ans;
	while(now != "-1"){
		ans.push_back({now,value[now]});
		now = ne[now];
	}
	// for(int i = 0; i < ans.size() ; i++){
		// cout<<ans[i].first<<" "<<ans[i].second<<endl;
	// }
	int l = 0,r = ans.size()-1;
	vector<pair<string,int>> res;
	while(l <= r){
		if(l == r){
			res.push_back({ans[l].first,ans[l].second});
			l++;
			r--;
		}else{
			res.push_back({ans[r].first,ans[r].second});
			res.push_back({ans[l].first,ans[l].second});
			l++;
			r--;
		}
	}
	int len = res.size();
	res.push_back({"-1",-1});
	for(int i = 0 ; i < len ; i++){
		cout<<res[i].first<<" "<<res[i].second<<" "<<res[i+1].first<<endl;
	}
}
signed main() {
    // ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int tt = 1;
    //    cin >> tt;
    while (tt--) {
        solve();
    }
    return 0;
}