Question

如何找到三个阵列中哪个阵列最长（最高计数）？

背景：

我有一个匹配的函数运行良好 - 三个字典，其中布尔值包含用户首选项，一篇文章有三个标记类别，函数检查标记A在字典A中打开，标记B在字典B中打开等等

现在要求标签A中有N个条目，标签B中有N个条目等等

因此，三个标签阵列中的每一个都可以有不同的长度，我能想到的最简单的方法是从ArrayA，ArrayB和ArrayC找到最长的数组（包含大多数条目）

这是我原来的工作循环

for (id myArrayElement in storyArray) {

    NSString *myString = [NSString stringWithString:[myArrayElement industryA]];
    NSString *myIssue = [NSString stringWithString:[myArrayElement issueA]];
    NSString *myService = [NSString stringWithString:[myArrayElement serviceA]];

    if (
        [prefsDictionary valueForKeyPath:[NSString stringWithFormat:@"Industries.%@", myString]] || 
        [prefsDictionary valueForKeyPath:[NSString stringWithFormat:@"Issues.%@", myIssueElement]] || 
        [prefsDictionary valueForKeyPath:[NSString stringWithFormat:@"Services.%@", myService]]
        ) {

        // One of the story's tags matches a key in one of the corresponding dictionaries
        // Look up what this preference is set to

        NSString *keyvalue = [[prefsDictionary valueForKey:@"Industries"] valueForKey:myString];
        NSString *Issuesvalue = [[prefsDictionary valueForKey:@"Issues"] valueForKey:myIssueElement];
        NSString *Servicevalue = [[prefsDictionary valueForKey:@"Services"] valueForKey:myService];

        if (
            [keyvalue isEqualToString:@"1"] || 
            [Issuesvalue isEqualToString:@"1"] || 
            [Servicevalue isEqualToString:@"1"]
            ) {

            // It's a match, add the story
            [self.favList addObject:myArrayElement];
        }

    } // prefsDictionary End if

我正在考虑这样做的最好方法，其中三个输入可以是任意长度的数组

for (id myArrayElement in delegate.storyArray) {

    NSArray *industyArr = [[myArrayElement industryA] componentsSeparatedByString:@"|"];
    NSArray *issueArr = [[myArrayElement issueA] componentsSeparatedByString:@"|"];
    NSArray *serviceArr = [[myArrayElement serviceA] componentsSeparatedByString:@"|"];

    // We need to find longest array
    // Pad the shorter arrays, or use if ([array count] >= 4) {id obj = [scores objectAtIndex:3];}
    // Then loop using the largest array length

    for (loop longest array length) {

               // get nth entry in industyArr... thisIndustry
               // get nth entry in issueArr...   thisIssue
               // get nth entry in serviceArr... thisService

        if (
            [prefsDictionary valueForKeyPath:[NSString stringWithFormat:@"Industries.%@", thisIndustry]] || 
            [prefsDictionary valueForKeyPath:[NSString stringWithFormat:@"Issues.%@", thisIssue]] || 
            [prefsDictionary valueForKeyPath:[NSString stringWithFormat:@"Services.%@", thisService]]
            ) {

            // One of the story's tags matches a key in one of the corresponding dictionaries

            NSString *keyvalue = [[prefsDictionary valueForKey:@"Industries"] valueForKey:thisIndustry];
            NSString *Issuesvalue = [[prefsDictionary valueForKey:@"Issues"] valueForKey:thisIssue];
            NSString *Servicevalue = [[prefsDictionary valueForKey:@"Services"] valueForKey:thisService];

            if (
                [keyvalue isEqualToString:@"1"] || 
                [Issuesvalue isEqualToString:@"1"] || 
                [Servicevalue isEqualToString:@"1"]
                ) {

                // It's a match, add the story
                [self.favList addObject:myArrayElement];

                // EXIT THE INNER LOOP NOW WE HAVE A MATCH
            }
        } // prefsDictionary End if
    } // End myIssueElement for
} // End myArrayElement for

除非有人有一个很棒的主意......

Answer 1

如果您只是想确保查看每个数组中的所有值，我实际上会支持nielsbot。只需插入

MAX(arrayA.count, MAX(arrayB.count, arrayC.count))

进入你的for循环参数应覆盖它。

for( int i=0; i < MAX(arrayA.count, MAX(arrayB.count, arrayC.count)); i++ ) {
    // Blah Blah Blah
}

MAX（）返回两个值中较大的一个，使您可以轻松选出最高计数。从您的示例代码中，您似乎并不需要最长的数组，只需要最高数量。

Answer 2

如果您要做的就是找出哪个数组的计数最高，那么这就是我要采取的方法：

NSArray *largestArray = arrayA;
if ([largestArray count] < [arrayB count]) {
    largestArray = arrayB;
}
if ([largestArray count] < [arrayC count]) {
    largestArray = arrayC;
}

如果有更多的阵列，这不是最佳方法，但只有三个，它应该做得很好。这应该每次返回最大的数组

查找具有最高计数的阵列

2 个答案: