Ios 算法：解析时间

我有一个需要在固定时间内解决的问题（关键在4小时内，重要在10小时内，正常在24小时内）。如果中间有非工作时间或假期，查询时间可以增加。非工作时间可以是一整天，也可以是一天中的某个小时

例如：周一上午8:02提出的关键问题应在下午1:00前得到解决

//Working hours of week 
Mon  09:00am - 01:00pm
Tue  Holiday
Wed  09:00am - 05:00pm(non working between 01:00pm - 02:00pm)
Thu  11:00am - 03:00pm(non working between 01:00pm - 02:00pm)
Fri  09:00am - 05:00pm(non working between 01:00pm - 02:00pm)

输入为查询类型和查询日志时间。需要的输出是解析时间

我的方法：

将所有工时添加到查询中，如同所有工时都在工作一样。因此，如果关键enq的记录时间为上午08:02，则添加4小时，即下午12:02作为恢复时间

输入while（true）循环，检查下一个hr是否为工作时间。如果是，跳过该hr，否则将其添加到res time并继续

但是，当日志时间不起作用时，这种方法不会给我正确的结果

NSDate *enqExpiration = nil;
    int hoursNeededToCoverEnq = 0;
    switch (enqType) {
        case normalPriority:
            hoursNeededToCoverEnq = lowResHours;
            enqExpiration = [NSDate dateWithTimeInterval:lowResHours*hour sinceDate:capturedTime];
            break;
        case importantPriority:
            hoursNeededToCoverEnq = mediumResHours;
            enqExpiration = [NSDate dateWithTimeInterval:mediumResHours*hour sinceDate:capturedTime];
            break;
        case criticalPriority:
            hoursNeededToCoverEnq = highResHours;
            enqExpiration = [NSDate dateWithTimeInterval:highResHours*hour sinceDate:capturedTime];
            break;
    }

    int aggregatedHrs = 0;

    while(true){
        capturedTime = [NSDate dateWithTimeInterval:1*hour sinceDate:capturedTime];

        NSDateFormatter *dayFormat = [[NSDateFormatter alloc] init];
        [dayFormat setTimeZone: [NSTimeZone timeZoneWithAbbreviation:@"GMT"]];
        [dayFormat setDateFormat:@"EEEE"];

        NSString* capturedDay = [dayFormat stringFromDate:capturedTime];
        [dayFormat setDateFormat:@"dd/MM/yyyy"];

        NSString* todayDate = [dayFormat stringFromDate:capturedTime];

        if([weekends containsObject:capturedDay]){
            enqExpiration = [NSDate dateWithTimeInterval:1*hour sinceDate:enqExpiration];
            continue;

        }else if([holidays containsObject:todayDate]){
            enqExpiration = [NSDate dateWithTimeInterval:1*hour sinceDate:enqExpiration];
            continue;

        }else{
            NSString *openTime, *closeTime;
            for(int i=0; i<weekdaysTime.count; i++){
                if([[((NSMutableDictionary*)[weekdaysTime objectAtIndex:i]) valueForKey:@"workingDay"] isEqualToString:capturedDay]){
                    openTime = [((NSMutableDictionary*)[weekdaysTime objectAtIndex:i]) valueForKey:@"openingTime"];
                    closeTime = [((NSMutableDictionary*)[weekdaysTime objectAtIndex:i]) valueForKey:@"closingTime"];
                    break;
                }
            }
            NSDate *openTimeToday = nil, *closeTimeToday = nil;
            [dayFormat setDateFormat:@"dd/MM/yyyy hh:mmaa"];
            openTimeToday = [dayFormat dateFromString:[[todayDate stringByAppendingString:@" "] stringByAppendingString:openTime]];
            closeTimeToday = [dayFormat dateFromString:[[todayDate stringByAppendingString:@" "] stringByAppendingString:closeTime]];
            if([capturedTime compare:openTimeToday]==NSOrderedAscending || [capturedTime compare:closeTimeToday]==NSOrderedDescending ){
                enqExpiration = [NSDate dateWithTimeInterval:1*hour sinceDate:enqExpiration];
                continue;
            }
        }
        aggregatedHrs ++;
        if( aggregatedHrs == hoursNeededToCoverEnq )
            break;
    }
return enqExpiration;

NSDate*enqexpntiation=nil；
int HOURSENDETOCOVERENQ=0；
开关（enqType）{
案件优先权：
Hours needToCoverenq=低恢复小时数；
EnqExption=[NSDate dateWithTimeInterval:lowResHours*Hours sinceDate:capturedTime]；
打破
案例重要性优先：
所需的小时数=中等小时数；
EnqExpation=[NSDate dateWithTimeInterval:mediumResHours*Hours sinceDate:capturedTime]；
打破
关键案例优先级：
Hours needtocoverenq=高恢复小时数；
EnqExption=[NSDate dateWithTimeInterval:highResHours*Hours sinceDate:capturedTime]；
打破
}
int aggregatedHrs=0；
while（true）{
capturedTime=[NSDate DATE WITH TIME INTERVAL:1*HOURE sinceDate:capturedTime]；
NSDateFormatter*dayFormat=[[NSDateFormatter alloc]init]；
[dayFormat setTimeZone:[NSTimeZone TimeZoneWithabRevision:@“GMT”]；
[dayFormat setDateFormat:@“EEEE”]；
NSString*capturedDay=[dayFormat stringFromDate:capturedTime]；
[dayFormat setDateFormat:@“dd/MM/yyyy”]；
NSString*todayDate=[dayFormat stringFromDate:capturedTime]；
if（[sobject:capturedDay]）{
enqExpiration=[NSDate DateWithTime Interval:1*小时，自日期：enqExpiration]；
持续
}否则，如果（[假日包含对象：todayDate]）{
enqExpiration=[NSDate DateWithTime Interval:1*小时，自日期：enqExpiration]；
持续
}否则{
NSString*打开时间，*关闭时间；
对于（int i=0；i）（我认为
的规范需要在固定时间内解决
很少（而且语法上不确定）——“服务协议”详细说明了条款，您的问题就不会出现。（描述中的“日志时间”与代码中的
capturedTime
相同吗？）当日志时间不起作用时，
不会给我正确的结果
-那么将
capturedTime
设置为下一个工作小时（开始）如何？（我建议重新访问（并可能在最后链接，特别是“编写完美问题”和“如何以智能方式提问”））谢谢你，现在如果你能想出解决办法那就太好了！