Day04介绍了Reverse Polish Notation,主要是为了方便用栈这个结构计算数学表达式。复习下:
输入前:(3 + 2)*(4 + 6) 转化后:3 2 + 4 6 + * 结果:50
- (int)evaluateRPN:(NSString *)inputStr
{
DSStack *newStack = [[DSStack alloc] initWithSize:10];
for (int i =0 ; i<inputStr.length; i++)
{
unichar tempChar = [inputStr characterAtIndex:i];
if ([[NSString stringWithCharacters:&tempChar length:1] isEqualToString:@"*"])
{
NSNumber *a = [newStack popLastObject];
NSNumber *b = [newStack popLastObject];
[newStack push:[NSNumber numberWithInt:([a intValue]*[b intValue])]];
}
else if ([[NSString stringWithCharacters:&tempChar length:1] isEqualToString:@"/"])
{
NSNumber *a = [newStack popLastObject];
NSNumber *b = [newStack popLastObject];
[newStack push:[NSNumber numberWithInt:([a intValue]/[b intValue])]];
}
else if ([[NSString stringWithCharacters:&tempChar length:1] isEqualToString:@"+"])
{
NSNumber *a = [newStack popLastObject];
NSNumber *b = [newStack popLastObject];
[newStack push:[NSNumber numberWithInt:([a intValue]+[b intValue])]];
}
else if ([[NSString stringWithCharacters:&tempChar length:1] isEqualToString:@"-"])
{
NSNumber *a = [newStack popLastObject];
NSNumber *b = [newStack popLastObject];
[newStack push:[NSNumber numberWithInt:([a intValue]-[b intValue])]];
}
else
{
[newStack push:[NSNumber numberWithInt:[[NSString stringWithCharacters:&tempChar length:1] intValue]]];
}
}
return [[newStack popLastObject] intValue];
}
把中缀表达式转化为Reverse Polish Notation。
按照顺序进栈。
当进栈的元素是 ( + - * / )四个运算符,则弹出栈顶两个元素并计算,并且把结果入栈。
最终栈只剩下一个元素,这个元素就是最后的值。
GitHub链接[1]
本文由 投稿者 创作,文章地址:https://blog.isoyu.com/archives/100tianiosshujujiegouyusuanfashizhan-day05-zhandesuanfashizhan-evaluate-reverse-polish-nota.html
采用知识共享署名4.0 国际许可协议进行许可。除注明转载/出处外,均为本站原创或翻译,转载前请务必署名。最后编辑时间为:4月 1, 2019 at 07:41 下午