让我提出一个不同的算法依赖于一个查询,而不是通过一个数组的搜索。
设置:
迭代字典中的单词。对于每个单词,创建一个具有相同字符的字符串,按字母顺序排序。使用此字符串作为关键字,创建原始单词数组的字典。
用法:
现在你可以做任何字符组合的检查非常快:就像上面的字符排序和查找产生的向上键在地图。
例子:
原始数组:(bond, Mary, army)
字谜查找图:
{
bdno : (bond),
amry : (Mary, army),
}
使用这种地图是非常快的检查任何字的字谜。不需要迭代字典数组。
编辑:
我提出的算法分割在三个部分:
- 甲设置方法来构建从对象的字典中查找图:
anagramMap
- 的方法,来计算字符按字符排序的键:
anagramKey
- 一种算法,可查找包含在九个字母单词中的所有字符的排列并查找w地图上的ords:
findAnagrams
。
这里的所有三种方法作为一个类别的上NSString
实现:
@interface NSString (NSStringAnagramAdditions)
- (NSSet *)findAnagrams;
@end
@implementation NSString (NSStringAnagramAdditions)
+ (NSDictionary *)anagramMap
{
static NSDictionary *anagramMap;
if (anagramMap != nil)
return anagramMap;
// this file is present on Mac OS and other unix variants
NSString *allWords = [NSString stringWithContentsOfFile:@"/usr/share/dict/words"
encoding:NSUTF8StringEncoding
error:NULL];
NSMutableDictionary *map = [NSMutableDictionary dictionary];
@autoreleasepool {
[allWords enumerateLinesUsingBlock:^(NSString *word, BOOL *stop) {
NSString *key = [word anagramKey];
if (key == nil)
return;
NSMutableArray *keyWords = [map objectForKey:key];
if (keyWords == nil) {
keyWords = [NSMutableArray array];
[map setObject:keyWords forKey:key];
}
[keyWords addObject:word];
}];
}
anagramMap = map;
return anagramMap;
}
- (NSString *)anagramKey
{
NSString *lowercaseWord = [self lowercaseString];
// make sure to take the length *after* lowercase. it might change!
NSUInteger length = [lowercaseWord length];
// in this case we're only interested in anagrams 4 - 9 characters long
if (length < 4 || length > 9)
return nil;
unichar sortedWord[length];
[lowercaseWord getCharacters:sortedWord range:(NSRange){0, length}];
qsort_b(sortedWord, length, sizeof(unichar), ^int(const void *aPtr, const void *bPtr) {
int a = *(const unichar *)aPtr;
int b = *(const unichar *)bPtr;
return b - a;
});
return [NSString stringWithCharacters:sortedWord length:length];
}
- (NSSet *)findAnagrams
{
unichar nineCharacters[9];
NSString *anagramKey = [self anagramKey];
// make sure this word is not too long/short.
if (anagramKey == nil)
return nil;
[anagramKey getCharacters:nineCharacters range:(NSRange){0, 9}];
NSUInteger middleCharPos = [anagramKey rangeOfString:[self substringWithRange:(NSRange){4, 1}]].location;
NSMutableSet *anagrams = [NSMutableSet set];
// 0x1ff means first 9 bits set: one for each character
for (NSUInteger i = 0; i <= 0x1ff; i += 1) {
// skip permutations that do not contain the middle letter
if ((i & (1 << middleCharPos)) == 0)
continue;
NSUInteger length = 0;
unichar permutation[9];
for (int bit = 0; bit <= 9; bit += 1) {
if (i & (1 << bit)) {
permutation[length] = nineCharacters[bit];
length += 1;
}
}
if (length < 4)
continue;
NSString *permutationString = [NSString stringWithCharacters:permutation length:length];
NSArray *matchingAnagrams = [[self class] anagramMap][permutationString];
for (NSString *word in matchingAnagrams)
[anagrams addObject:word];
}
return anagrams;
}
@end
假设在一个名为变量测试字符串nineletters
你会使用日志的可能值:
for (NSString *anagram in [nineletters findAnagrams])
NSLog(@"%@", anagram);
我我面临同样的问题......但我没有从答案算法中获得所有可能的字典。 – 2013-10-10 08:06:30