在字符串中将两个字符的出现替换为彼此

Question

我遇到了一个问题，我需要在字符串中倒置两个字符。例如，如果我的字符串是"a * b/c“，并且我希望用/ and / with *替换出现的*。我希望得到的字符串是"a/b*c“。

使用stringByReplacingOccurrenceOfString:方法不起作用，因为我不希望第一轮替换影响第二轮：

string = @"a*b/c";
[string stringByReplacingOccurrencesOfString:@"*" withString:@"/"];
[string stringByReplacingOccurrencesOfString:@"/" withString:@"*"];

这就产生了"a*b*c"，这不是我想要的。有没有人知道一个有效的方法来实现这一点？

Answer 1

string = @"a*b/c";
[string stringByReplacingOccurrencesOfString:@"*" withString:@"&"];
[string stringByReplacingOccurrencesOfString:@"/" withString:@"*"];
[string stringByReplacingOccurrencesOfString:@"&" withString:@"/"];

Answer 2

使用NSScanner在字符串中移动，并在找到每个字符时替换它。这样，所有的换人都是在一次传球中完成的，你永远不会两次看到一个位置。

NSMutableString * fixedUpString = [NSMutableString string];
NSScanner * scanner = [NSScanner scannerWithString:origString];
NSCharacterSet * subCharacters = [NSCharacterSet characterSetWithCharactersInString:@"*/"];

while( ![scanner isAtEnd] ){

    // Pick up other characters.
    NSString * collector;
    if( [scanner scanUpToCharactersInSet:subCharacters intoString:&collector] ){
        [fixedUpString appendString:collector];
    }

    // This can easily be generalized with a loop over a mapping from
    // found characters to substitutions
    // Check which one we found
    if( [scanner scanString:@"*" intoString:nil] ){

        // Append the appropriate substitution.
        [fixedUpString appendString:@"/"];
    }
    else /* if( [scanner scanString:@"/" intoString:nil] ) */  {
        [fixedUpString appendString:@"*"];
    }
}

fixedUpString现在包含替代的内容。

正如我在评论中所指出的，这可以很容易地推广到任意数量的替换：

NSDictionary * substitutions = @{ @"a" : @"z", @"b" : @"y", ... };
NSCharacterSet * keyChars = [NSCharacterSet characterSetWithCharactersInString:[[substitutions allKeys] componentsJoinedByString:@""]];

...

// Check which one we found
for( NSString * keyChar in [substitutions allKeys] ){

    if( [scanner scanString:keyChar intoString:nil ){
        [fixedUpString appendString:substitutions[keyChar]];
        break;
    }
}

Answer 3

在不使用中间&字符的情况下，我不得不尝试这样做，虽然这当然更复杂，但这似乎也有效：

int main(int argc, char *argv[])
{

    NSString *s = @"1*2/3*4*5*6*7*8/2";

    NSArray *stars = [s componentsSeparatedByString:@"*"];
    NSMutableArray *slashes = [NSMutableArray array];
    for (NSString *star in stars)
    {
        [slashes addObject:[star componentsSeparatedByString:@"/"]];
    }

    NSMutableArray *newStars = [NSMutableArray array];
    for (NSArray *slash in slashes)
    {
        [newStars addObject:[slash componentsJoinedByString:@"*"]];
    }

    NSString *newString = [newStars componentsJoinedByString:@"/"];

    NSLog(@"%@", newString);

    return 0;
}

输出：

1/2*3/4/5/6/7/8*2