我在使用括号完成算法时遇到了一个令人沮丧的问题。我使用的数学库 DDMathParser 只处理以弧度表示的三角函数。如果想使用度数，他们必须调用dtor(deg_value). 问题是这增加了一个额外的括号，必须在最后考虑。
例如，数学表达式sin(cos(sin(x)))将sin(dtor(cos(dtor(sin(dtor(x)))在我的代码中转换为。但是，请注意，我需要两个额外的括号才能使其成为完整的数学表达式。因此，创建resolve_dtor().
这是我尝试的解决方案，想法是0表示左括号，用1表示左括号， dtor(并2表示右括号，从而完成0或1。
- (NSMutableString *)resolve_dtor:(NSMutableString *)exp
{
NSInteger mutable_length = [exp length];
NSMutableArray *paren_complete = [[NSMutableArray alloc] init]; // NO/YES
for (NSInteger index = 0; index < mutable_length; index++) {
if ([exp characterAtIndex:index] == '(') {
// Check if it is "dtor()"
if (index > 5 && [[exp substringWithRange:NSMakeRange(index - 4, 4)] isEqual:@"dtor"]) {
//dtor_array_index = [self find_incomplete_paren:paren_complete];
[paren_complete addObject:@1];
}
else
[paren_complete addObject:@0]; // 0 signifies an incomplete parenthetical expression
}
else if ([exp characterAtIndex:index] == ')' && [paren_complete count] >= 1) {
// Check if "dtor("
if (![self elem_is_zero:paren_complete]) {
// Add right-paren for "dtor("
[paren_complete replaceObjectAtIndex:[self find_incomplete_dtor:paren_complete] withObject:@2];
[exp insertString:@")" atIndex:index + 1];
mutable_length++;
index++;
}
else
[paren_complete replaceObjectAtIndex:[self find_incomplete_paren:paren_complete] withObject:@2];
}
else if ([paren_complete count] >= 1 && [[paren_complete objectAtIndex:0] isEqualToValue:@2]) {
// We know that everything is complete
[paren_complete removeAllObjects];
}
}
return exp;
}
- (bool)check_dtor:(NSMutableString *)exp
{
NSMutableArray *paren_complete = [[NSMutableArray alloc] init]; // NO/YES
for (NSInteger index = 0; index < [exp length]; index++) {
if ([exp characterAtIndex:index] == '(') {
// Check if it is "dtor()"
if (index > 5 && [[exp substringWithRange:NSMakeRange(index - 4, 4)] isEqual:@"dtor"]) {
//dtor_array_index = [self find_incomplete_paren:paren_complete];
[paren_complete addObject:@1];
}
else
[paren_complete addObject:@0]; // 0 signifies an incomplete parenthetical expression
}
else if ([exp characterAtIndex:index] == ')' && [paren_complete count] >= 1) {
// Check if "dtor("
if (![self elem_is_zero:paren_complete]) {
// Indicate "dtor(" at index is now complete
[paren_complete replaceObjectAtIndex:[self find_incomplete_dtor:paren_complete] withObject:@2];
}
else
[paren_complete replaceObjectAtIndex:[self find_incomplete_paren:paren_complete] withObject:@2];
}
else if ([paren_complete count] >= 1 && [[paren_complete objectAtIndex:0] isEqualToValue:@2]) {
// We know that everything is complete
[paren_complete removeAllObjects];
}
}
// Now step back and see if all the "dtor(" expressions are complete
for (NSInteger index = 0; index < [paren_complete count]; index++) {
if ([[paren_complete objectAtIndex:index] isEqualToValue:@0] || [[paren_complete objectAtIndex:index] isEqualToValue:@1]) {
return NO;
}
}
return YES;
}
似乎该算法适用于sin((3 + 3) + (6 - 3))（转换为sin(dtor((3 + 3) x (6 - 3))）但不适用于sin((3 + 3) + cos(3))（转换为sin(dtor((3 + 3) + cos(dtor(3)).
底线
这个半解决方案很可能过于复杂（似乎是我的常见问题之一），所以我想知道是否有更简单的方法可以做到这一点？
解决方案
这是我对他提供的@j_random_hacker 伪代码的解决方案：
- (NSMutableString *)resolve_dtor:(NSString *)exp
{
uint depth = 0;
NSMutableArray *stack = [[NSMutableArray alloc] init];
NSRegularExpression *regex_trig = [NSRegularExpression regularExpressionWithPattern:@"(sin|cos|tan|csc|sec|cot)" options:0 error:0];
NSRegularExpression *regex_trig2nd = [NSRegularExpression regularExpressionWithPattern:@"(asin|acos|atan|acsc|asec|acot)" options:0 error:0];
// need another regex for checking asin, etc. (because of differing index size)
NSMutableString *exp_copy = [NSMutableString stringWithString:exp];
for (NSInteger i = 0; i < [exp_copy length]; i++) {
// Check for it!
if ([exp_copy characterAtIndex:i] == '(') {
if (i >= 4) {
// check if i - 4
if ([regex_trig2nd numberOfMatchesInString:exp_copy options:0 range:NSMakeRange(i - 4, 4)] == 1) {
[stack addObject:@(depth)];
[exp_copy insertString:@"dtor(" atIndex:i + 1];
depth++;
}
}
else if (i >= 3) {
// check if i - 3
if ([regex_trig numberOfMatchesInString:exp_copy options:0 range:NSMakeRange(i - 3, 3)] == 1) {
[stack addObject:@(depth)];
[exp_copy insertString:@"dtor(" atIndex:i + 1];
depth++;
}
}
}
else if ([exp_copy characterAtIndex:i] == ')') {
depth--;
if ([stack count] > 0 && [[stack objectAtIndex:[stack count] - 1] isEqual: @(depth)]) {
[stack removeObjectAtIndex:[stack count] - 1];
[exp_copy insertString:@")" atIndex:i + 1];
}
}
}
return exp_copy;
}
有用！让我知道是否有任何小的更正可以添加，或者是否有更有效的方法。