1

如何优化这个嵌套的 for 循环？

程序应该遍历从单词文本文件创建的数组中的每个单词，如果大于 8 个字符，则将其添加到goodWords数组中。但需要注意的是，我只希望根词在 goodWords 数组中，例如：

如果 greet 被添加到数组中，我不想要问候语或问候语或问候语等。

    NSString *string = [NSString stringWithContentsOfFile:@"/Users/james/dev/WordParser/word.txt" encoding:NSUTF8StringEncoding error:NULL];
    NSArray *words = [string componentsSeparatedByString:@"\r\n"];
    NSMutableArray *goodWords = [NSMutableArray array];
    BOOL shouldAddToGoodWords = YES;

    for (NSString *word in words)
    {
        NSLog(@"Word: %@", word);

        if ([word length] > 8)
        {
            NSLog(@"Word is greater than 8");

            for (NSString *existingWord in [goodWords reverseObjectEnumerator])
            {
                NSLog(@"Existing Word: %@", existingWord);
                if ([word rangeOfString:existingWord].location != NSNotFound)
                {
                    NSLog(@"Not adding...");
                    shouldAddToGoodWords = NO;
                    break;
                }
            }

            if (shouldAddToGoodWords)
            {
                NSLog(@"Adding word: %@", word);
                [goodWords addObject:word];
            }
        }

        shouldAddToGoodWords = YES;
    }
4

3 回答 3

3

How about something like this?

//load the words from wherever
NSString * allWords = [NSString stringWithContentsOfFile:@"/usr/share/dict/words"];
//create a mutable array of the words
NSMutableArray * words = [[allWords componentsSeparatedByCharactersInSet:[NSCharacterSet newlineCharacterSet]] mutableCopy];
//remove any words that are shorter than 8 characters
[words filterUsingPredicate:[NSPredicate predicateWithFormat:@"length >= 8"]];
//sort the words in ascending order
[words sortUsingSelector:@selector(caseInsensitiveCompare:)];

//create a set of indexes (these will be the non-root words)
NSMutableIndexSet * badIndexes = [NSMutableIndexSet indexSet];
//remember our current root word
NSString * currentRoot = nil;
NSUInteger count = [words count];
//loop through the words
for (NSUInteger i = 0; i < count; ++i) {
    NSString * word = [words objectAtIndex:i];
    if (currentRoot == nil) {
        //base case
        currentRoot = word;
    } else if ([word hasPrefix:currentRoot]) {
        //word is a non-root word.  remember this index to remove it later
        [badIndexes addIndex:i];
    } else {
        //no match. this word is our new root
        currentRoot = word;
    }
}
//remove the non-root words
[words removeObjectsAtIndexes:badIndexes];
NSLog(@"%@", words);
[words release];

This runs very very quickly on my machine (2.8GHz MBP).

于 2010-08-06T20:32:06.350 回答
2

Trie似乎适合您的目的。它就像一个哈希，对于检测给定的字符串是否是已经看到的字符串的前缀很有用。

于 2010-08-06T18:59:28.130 回答
1

I used an NSSet to ensure that you only have 1 copy of a word added at a time. It will add a word if the NSSet does not already contain it. It then checks to see if the new word is a substring for any word that has already been added, if true then it won't add the new word. It's case-insensitive as well.

What I've written is a refactoring of your code. It's probably not that much faster but you really do want a tree data structure if you want to make it a lot faster when you want to search for words that have already been added to your tree.

Take a look at RedBlack Trees or B-Trees.

Words.txt

objective
objectively
cappucin
cappucino
cappucine
programme
programmer
programmatic
programmatically

Source Code

- (void)addRootWords {

    NSString        *textFile = [[NSBundle mainBundle] pathForResource:@"words" ofType:@"txt"];
    NSString        *string = [NSString stringWithContentsOfFile:textFile encoding:NSUTF8StringEncoding error:NULL];
    NSArray         *wordFile = [string componentsSeparatedByString:@"\n"];
    NSMutableSet    *goodWords = [[NSMutableSet alloc] init];

    for (NSString *newWord in wordFile)
    {
        NSLog(@"Word: %@", newWord);
        if ([newWord length] > 8)
        {
            NSLog(@"Word '%@' contains 8 or more characters", newWord);
            BOOL shouldAddWord = NO;
            if ( [goodWords containsObject:newWord] == NO) {
                shouldAddWord = YES;
            }
            for (NSString *existingWord in goodWords)
            {
                NSRange textRange = [[newWord lowercaseString] rangeOfString:[existingWord lowercaseString]];
                if( textRange.location != NSNotFound ) {
                    // newWord contains the a substring of existingWord
                    shouldAddWord = NO;
                    break;
                }
                NSLog(@"(word:%@) does not contain (substring:%@)", newWord, existingWord);
                shouldAddWord = YES;
            }
            if (shouldAddWord) {
                NSLog(@"Adding word: %@", newWord);
                [goodWords addObject:newWord];
            }
        }
    }

    NSLog(@"***Added words***");
    int count = 1;
    for (NSString *word in goodWords) {
        NSLog(@"%d: %@", count, word);
        count++;
    }

    [goodWords release];
}

Output:

***Added words***
1: cappucino
2: programme
3: objective
4: programmatic
5: cappucine
于 2010-08-06T20:03:49.517 回答