I read about Big-O Notation from here and had few questions on calculating the complexity.So for the below code i have calculated the complexity. need your inputs for the same.
private void reverse(String strToRevers)
{
if(strToRevers.length() == 0)
{
return ;
}
else
{
reverse(strToRevers.substring(1));
System.out.print(strToRevers.charAt(0));
}
}
If the memory factor is considered then the complexity of above code for a string of n characters is O(n^2). The explanation is for a string that consists of n characters, the below function would be called recursively n-1 times and each function call c开发者_运维百科reates a string of single character(stringToReverse.charAT(0)). Hence it is n*(n-1)*2 which translates to o(n^2). Let me know if this is right ?
Hence it is n*(n-1)*2 which translates to o(n^2). Let me know if this is right ?
Almost: it's n * (n-1) / 2
, not *2
, which is also O(n^2). Note that o(n^2) (little-O) means something else, so the distinction is important.
This is assuming we're considering this as pseudocode. Language-specific implementations and smart compilers may be able to improve the running time substantially. For instance, a compiler that can observe that you're simply reversing the string might just do an in-place reverse, which is O(n).
Looks like Java, so it's not O(n**2). That's because strings share the underlying character sequence buffers; they can do this because they are immutable objects.
But it is O(n) in stack space. That's not good. It's better to allocate a mutable working buffer of characters, reverse the string into that, and then print the whole lot at once.
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