創建的NSMutableString我有兩個字符串:從另外兩個NSMutableStrings
@"--U"
和@"-O-"
並希望創建另一個的NSMutableString,使用兩個已知條件,使@"-OU"
。有誰知道我該怎麼做?
創建的NSMutableString我有兩個字符串:從另外兩個NSMutableStrings
@"--U"
和@"-O-"
並希望創建另一個的NSMutableString,使用兩個已知條件,使@"-OU"
。有誰知道我該怎麼做?
注意,下面的代碼假設S1和S2具有相同的長度,否則會拋出在某些時候例外,所以做檢查:)
- (NSMutableString *)concatString:(NSString *)s1 withString:(NSString *)s2
{
NSMutableString *result = [NSMutableString stringWithCapacity:[s1 length]];
for (int i = 0; i < [s1 length]; i++) {
unichar c = [s1 characterAtIndex:i];
if (c != '-') {
[result appendFormat:@"%c", c];
}
else {
[result appendFormat:@"%c", [s2 characterAtIndex:i]];
}
}
return result;
}
NSString *[email protected]"-0-";
NSString *[email protected]"--U";
NSString *temp1=[t1 substringWithRange:NSMakeRange(0, 2)];
NSString *temp2=[t2 substringFromIndex:2];
NSLog(@"%@",[NSString stringWithFormat:@"%@%@",temp1,temp2]);
這個版本有點比Nick更囉嗦,但把它分解成C函數和尾遞歸,所以它可能運行得更快。它還處理不同長度的字符串,選擇鏡像較短的字符串長度。
注:我還沒有運行此代碼,所以它可能是越野車或缺少明顯的東西。
void recursiveStringMerge(unichar* string1, unichar* string2, unichar* result) {
if (string1[0] == '\0' || string2[0] == '\0') {
result[0] = '\0'; //properly end the string
return; //no use in trying to add more to this string
}
else if (string1[0] != '-') {
result[0] = string1[0];
}
else {
result[0] = string2[0];
}
//move on to the next unichar in each array
recursiveStringMerge(string1+1, string2+1, result+1);
}
- (NSMutableString *)concatString:(NSString *)s1 withString:(NSString *)s2 {
NSUInteger resultLength;
NSUInteger s1Length = [s1 length]+1; //ensure space for NULL with the +1
NSUInteger s2Length = [s2 length]+1;
resultLength = (s1Length <= s2Length) ? s1Length : s2Length; //only need the shortest
unichar* result = malloc(resultLength*sizeof(unichar));
unichar *string1 = calloc(s1Length, sizeof(unichar));
[s1 getCharacters:buffer];
unichar *string2 = calloc(s2Length, sizeof(unichar));
[s2 getCharacters:buffer];
recursiveStringMerge(string1, string2, result);
return [NSString stringWithCharacters: result length: resultLength];
}
這是完美的謝謝。訣竅是使用unichar :) – locoboy 2011-04-21 05:13:03
還有一個關於代碼的假設:如果s1和s2在同一位置都有非「 - 」字符,則來自s1的字符將優先:) – Nick 2011-04-21 05:57:37