Problem-solving ideas:
1. 读入数据，将其截取转化为时间字符串
2. 将每个时间字符串，对应的时分秒转化为整数，并存入数组
3. 如果时间字符串以（+1）结尾，则h+24,以（+2）结尾，则h+48
4. 算出两次 起飞和降落的时间差，求出其平均值，就是结果
5. 最后答案，需要将结果中 个位数前面加上零
import java.util.ArrayList;
import java.util.Scanner;
public class 航班时间 {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in = new Scanner(System.in);
int N = Integer.parseInt(in.nextLine());
ArrayList<String> results = new ArrayList<String>();
while(N-->0) {
String str1 = in.nextLine();
String str2 = in.nextLine();
String[] str0 = {
str1.substring(0, 8),str1.substring(9),str2.substring(0, 8),str2.substring(9)};
// 将时间转成int数组
int[] time1 = time_arr(str0[0]);
int[] time2 = time_arr(str0[1]);
int[] time3 = time_arr(str0[2]);
int[] time4 = time_arr(str0[3]);
//求时间差
int[] result1 = subtime(time1,time2);
int[] result2 = subtime(time3,time4);
//算出两次航行时间差的平均值
int [] result = aver_time(result1,result2);
//对时间进行补0格式化
String ans[] = Zero_ans(result);
//输出
results.add(ans[0]+":"+ans[1]+":"+ans[2]);
}
for(int i=0;i<results.size();i++) {
System.out.println(results.get(i));
}
}
static String[] Zero_ans(int[] result) {
String h,m,s;
if(result[0]<10) {
h="0"+result[0];
}else {
h=""+result[0];
}
if(result[1]<10) {
m="0"+result[1];
}else {
m=""+result[1];
}
if(result[2]<10) {
s="0"+result[2];
}else {
s=""+result[2];
}
return new String[] {
h,m,s};
}
static int[] aver_time(int[] result1, int[] result2) {
int h,m,s;
if((result1[0]+result2[0])%2==0) {
h = (result1[0]+result2[0])/2;
}else {
result1[1]+=60;
h=(result1[0]+result2[0])%2;
}
if((result1[1]+result2[1])%2==0) {
m = (result1[1]+result2[1])/2;
}else {
result1[2]+=60;
m=(result1[1]+result2[1])%2;
}
s = (result1[2]+result2[2])/2;
return new int[] {
h,m,s};
}
private static int[] subtime(int[] time1, int[] time2) {
int h,m,s;
if(time2[2]>=time1[2]) {
s = time2[2]-time1[2];
}else {
time2[1]--;
s = 60+time2[2]-time1[2];
}
if(time2[1]>=time1[1]) {
m = time2[1]-time1[1];
}else {
time2[0]--;
m = 60+time2[1]-time1[1];
}
h = time2[0]-time1[0];
return new int[] {
h,m,s};
}
static int[] time_arr(String string) {
int h,m,s;
if(string.contains("(+1)")) {
h = 24+Integer.parseInt(string.substring(0, 2));
}else if(string.contains("(+2)")) {
h = 48+Integer.parseInt(string.substring(0, 2));
}else {
h = Integer.parseInt(string.substring(0, 2));
}
m = Integer.parseInt(string.substring(3, 5));
s = Integer.parseInt(string.substring(6,8));
return new int[] {
h,m,s};
}
}
Reference: https://blog.csdn.net/weixin_44107920/article/details/108029975