题解 | #合并表记录#

import java.util.*;

public class Main{
    public static void main(String[] args){
        Scanner scanner1 = new Scanner(System.in);
        int n = Integer.valueOf(scanner1.nextLine());
        int len = 0;
        if(n < 1 || n > 500) return;
        Map<Integer,Integer> map = new HashMap<>();
        while(len < n && scanner1.hasNext()) {
            String[] strArray = scanner1.nextLine().split(" ");
            if(strArray.length > 2) {
                continue;
            }
            Integer one = Integer.valueOf(strArray[0]);
            Integer two = Integer.valueOf(strArray[1]);
            if(map.containsKey(one)){
                map.put(one,map.get(one) + two);
            }else{
                map.put(one,two);
            }
            len++;
        }
        if(len==n) {
            LinkedHashMap<Integer,Integer> res = ConTableRec(map);
            for(Map.Entry<Integer,Integer> m:res.entrySet()){
                System.out.println(m.getKey() +" " + m.getValue());
            }

        }
    }
    
    private static LinkedHashMap<Integer,Integer> ConTableRec(Map<Integer,Integer> maps){
        LinkedHashMap<Integer,Integer> map = new LinkedHashMap<>();
        //定义一个key数组
        int[] keyArray = new int[maps.keySet().size()];
        Iterator<Map.Entry<Integer,Integer>> it = maps.entrySet().iterator();
        int i = 0;
        while(it.hasNext() && i<keyArray.length){
           Map.Entry<Integer,Integer> entry = it.next();
            keyArray[i] = entry.getKey();
            i++;
        }
        //对数组进行升序排序
        for(int s = 0;s < keyArray.length;s++){
            for(int j = keyArray.length - 1;j > s;j--){
                if(keyArray[s]>keyArray[j]){
                    int tmp = keyArray[j];
                    keyArray[j] = keyArray[s];
                    keyArray[s] = tmp;
                }
            }
        }
        //对map遍历赋值
        for(int t = 0;t < keyArray.length;t++){
            map.put(keyArray[t],maps.get(keyArray[t]));
        }
        return map;
    }
}