java 数组去重复

{"deviceid":"460706130100567","distance":88.6,"maxspeed":85.0,"runningtime":"1时22分3秒"},
{"deviceid":"460706130100567","distance":0.0,"maxspeed":0.0,"runningtime":"0秒"},
{"deviceid":"460706130100567","distance":55.6,"maxspeed":82.0,"runningtime":"59分14秒"},
{"deviceid":"460706130100567","distance":0.0,"maxspeed":0.0,"runningtime":"0秒"},
{"deviceid":"460706130100567","distance":0.0,"maxspeed":0.0,"runningtime":"0秒"}
把上面数据写成一个
{"deviceid":"460706130100567","distance":144.2,"maxspeed":85.0,"runningtime":"2时21分17秒"}
这个460706130100567只取一个，怎么弄？？？？？？

你这个数据{"deviceid":"460706130100567","distance":88.6,"maxspeed":85.0,"runningtime":"1时22分3秒"},是json格式的吧，要转化才方便处理，或者换种格式。

我写个大概，
public class CombinTest {
//求距离之和
static double calDistance(String[] str){
double dl=0;
for(int i=0;i dl=dl+Double.parseDouble(str[i].split(":")[1]);
}
return dl;
}
//求最大值
static double maxNum(double[] dl){
double max=0;
for(int i=0;i if(dl[i]>max){
max=dl[i];
}
}
return max;
}
//求最大速度
static double maxSpeed(String[] str){
double[] tempArray=new double[str.length];
for(int i=0;i tempArray[i]=Double.parseDouble(str[i].split(":")[1]);
}
return maxNum(tempArray);
}

//求时间之和
static String calTime(String[] str){
int sumSecond=0;
for(int i=0;i sumSecond=sumSecond+dealTime(str[i].split(":")[1]);
}
String strtemp=(sumSecond/3600)+"时"+((sumSecond%3600)/60)+"分"+((sumSecond%3600)%60)+"秒";
return strtemp;
}

//处理时间
static int dealTime(String str){
int sumSecond=0;
if(!(str.indexOf("时")==-1)){
int hour=Integer.parseInt(str.split("时")[0]);
int minute=Integer.parseInt(str.split("时")[1].split("分")[0]);
int second=Integer.parseInt(str.split("时")[1].split("分")[0].split("秒")[0]);
sumSecond=hour*3600+minute*60+second;
}else if(!(str.indexOf("分")==-1)){
int minute=Integer.parseInt(str.split("分")[0]);
int second=Integer.parseInt(str.split("秒")[0]);
sumSecond=minute*60+second;
}else{
int second=Integer.parseInt(str.split("秒")[0]);
sumSecond=second;
}
return sumSecond;
}

public static void main(String[] args){

}
}

是不是可以考虑用map集合做

又提了一次，你去看看我在你另一个问题里的回答

list array2 = new arraylist(); for (int i = 0; i < array.length; i++) { boolean exist = false; for (int j = 0; j < array2.size(); j++) { if (arrays.equals(array[i], array2.get(j))) { exist = true; break; } } if (!exist) array2.add(array[i]); } string[][] x = array2.toarray(new string[0][]);

我觉得这个做4个数组比较好分别是: divices,distances,msds,rts 每个数组互相对应 如果 divices 前5个相等 则 diviceid 取 divices[4] 后面相应取前5个作运算