Question

我试图在NSString中重复使用他们的计数字符。

例如，setupsetpreset作为输入，它返回＆＃39; set＆＃39;以及计数3.

让我用另一个例子来解释。

输入1： upsetpreset

预期输出1：设置2

输入2： qwertygdgpopgdg

预期输出2： gdg 2

Answer 1

请参阅下文

NSString *string     = @"This is a test. This is only a test.";
NSCountedSet *countedSet = [NSCountedSet new];

[string enumerateSubstringsInRange:NSMakeRange(0, [string length])
                       options:NSStringEnumerationByWords | NSStringEnumerationLocalized
                    usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop){

                        // This block is called once for each word in the string.
                        [countedSet addObject:substring];

                        // If you want to ignore case, so that "this" and "This" 
                        // are counted the same, use this line instead to convert
                        // each word to lowercase first:
                        // [countedSet addObject:[substring lowercaseString]];
                    }];

NSLog(@"%@", countedSet);

结果：

“这个” - 2 “是” - 2 “a2 “测试” - 2 “只” - 1

Answer 2

在获得大量自定义逻辑以获取非空格字符串中的常用字符后，我得到了解决方案： -

NSString *string=@"setupsetpreset";
string=@"sheraheshehehesheshe";
NSMutableArray *a=[[NSMutableArray alloc] init];
NSMutableArray *b=[[NSMutableArray alloc] init];
for (int i=0; i<(string.length); i++) {
    [a addObject:[NSString stringWithFormat:@"%c",[string characterAtIndex:i]]];
}
NSCountedSet *countedSet = [NSCountedSet new];
[string enumerateSubstringsInRange:NSMakeRange(0, [string length])
                           options:NSStringEnumerationByComposedCharacterSequences
                        usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop){
                            [countedSet addObject:substring];
                        }];
NSMutableArray *c=[[NSMutableArray alloc] init];
for (int i=0; i<(string.length); i++) {
    if([countedSet countForObject:[a objectAtIndex:i]]>=2){
        if(![b containsObject:[a objectAtIndex:i]]){
            [c addObject:[NSString stringWithFormat:@"%d",(int)[countedSet countForObject:[a objectAtIndex:i]]]];
        }
        [b addObject:[a objectAtIndex:i]];
    }
}

int total=0;
for (NSString *s in c){
    int m=[s intValue];
    total+=m;
}
total=total/c.count;
NSString *s=[b componentsJoinedByString:@""];
[c removeAllObjects];
for (int i=0; i<(s.length); i++) {
    if([countedSet countForObject:[b objectAtIndex:i]]>=total){

        [c addObject:[b objectAtIndex:i]];
    }
}
s=[c componentsJoinedByString:@""];
int iterate=0;
NSString  *substring;
NSMutableDictionary *dict=[[NSMutableDictionary alloc] init];
while (iterate<s.length-total) {
    NSString *searchstring=[s substringWithRange:NSMakeRange(iterate, total)];
    if ([string rangeOfString:searchstring].location == NSNotFound) {
    } else {
            [dict setObject:[NSNumber numberWithInt:[[dict objectForKey:searchstring] intValue]+1] forKey:searchstring];
        }
        iterate++;
    }
    [a removeAllObjects];
    [b removeAllObjects];
    a=[dict.allValues mutableCopy];
    b=[dict.allKeys mutableCopy];
    int max = [[a valueForKeyPath:@"@max.intValue"] intValue];
    NSInteger Aindex = [a indexOfObject:[NSNumber numberWithInt:max]];
    substring =[NSString stringWithFormat:@"%@",[b objectAtIndex:Aindex]];
    NSLog(@"Substring %@ is repeated %d times",substring,max);

希望它有助于满足某人的需求:)感谢Kiran为我提供了一组逻辑，这些逻辑是您的查询所必需的部分和Ruchira。

从NSString获取公共字符作为子字符串

2 个答案: