The function has several drawbacks.
For starters it should be declared like
std::string::size_type strStr( const std::string &haystack, const std::string &needle );
and if the second string is not found in the first string the function should return std::string::npos as all similar member functions of the class std::string do.
The function parameters shell be of constant referenced types.
The condition in this if-statement
if (haystack.empty() && !needle.empty())
has a redundant operand. It could be rewritten like
if (haystack.empty())
This loop
for (i; i < haystack.length(); i++)
should stop its iterations when the size of the tail of the first string is less than the size of the second string.
in this if-statement
if (haystack[i++] != needle[j]) {
the variable i is incremented that results in incrementing the variable two times: one in this statement and the second time in the loop.
The second pair of these statements
if (j == needle.length()) {
return ans;
is redundant.
The function can be written the following way as it is shown in the demonstrative program.
#include <iostream>
#include <string>
std::string::size_type strStr( const std::string &haystack, const std::string &needle )
{
if ( needle.empty() )
{
return 0;
}
else if ( haystack.empty() )
{
return -std::string::npos;
}
else
{
std::string::size_type ans = std::string::npos;
auto n1 = haystack.length();
auto n2 = needle.length();
for ( std::string::size_type i = 0; ans == std::string::npos && i + n2 <= n1; i++ )
{
std::string::size_type j = 0;
while ( j < n2 && haystack[i+j] == needle[j] ) j++;
if ( j == n2 ) ans = i;
}
return ans;
}
}
int main()
{
std::string haystack( "mississippi" );
std::string needle( "issip" );
std::cout << strStr( haystack, needle ) << '\n';
return 0;
}
Its output is
4
Answer from Vlad from Moscow on Stack OverflowThe function has several drawbacks.
For starters it should be declared like
std::string::size_type strStr( const std::string &haystack, const std::string &needle );
and if the second string is not found in the first string the function should return std::string::npos as all similar member functions of the class std::string do.
The function parameters shell be of constant referenced types.
The condition in this if-statement
if (haystack.empty() && !needle.empty())
has a redundant operand. It could be rewritten like
if (haystack.empty())
This loop
for (i; i < haystack.length(); i++)
should stop its iterations when the size of the tail of the first string is less than the size of the second string.
in this if-statement
if (haystack[i++] != needle[j]) {
the variable i is incremented that results in incrementing the variable two times: one in this statement and the second time in the loop.
The second pair of these statements
if (j == needle.length()) {
return ans;
is redundant.
The function can be written the following way as it is shown in the demonstrative program.
#include <iostream>
#include <string>
std::string::size_type strStr( const std::string &haystack, const std::string &needle )
{
if ( needle.empty() )
{
return 0;
}
else if ( haystack.empty() )
{
return -std::string::npos;
}
else
{
std::string::size_type ans = std::string::npos;
auto n1 = haystack.length();
auto n2 = needle.length();
for ( std::string::size_type i = 0; ans == std::string::npos && i + n2 <= n1; i++ )
{
std::string::size_type j = 0;
while ( j < n2 && haystack[i+j] == needle[j] ) j++;
if ( j == n2 ) ans = i;
}
return ans;
}
}
int main()
{
std::string haystack( "mississippi" );
std::string needle( "issip" );
std::cout << strStr( haystack, needle ) << '\n';
return 0;
}
Its output is
4
The problem is that you modify i in
if (haystack[i++] != needle[j]) {
Thus preventing a second potential match from being explored. Try
if (haystack[i + j] != needle[j]) {
and fix any knock-on issues. I expect it to work as-is, though.
Videos
programming challenge - LeetCode:implement strstr C# - Code Review Stack Exchange
debugging - Leetcode 28. Implement strStr() -- Why my code failed? - Stack Overflow
LeetCode #28 - Implement StrStr (Easy)
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More on reddit.comAs dfhwze and Roland already pointed out, a hash alone is not sufficient to determine whether two things are equal, so you still need to do a string comparison afterwards if the hashes match. Otherwise you will get wrong results from time to time. Not to mention the effect of hash randomization between different application runs...
The idea behind Rabin-Karp's use of hashes is to replace costly string comparisons with cheap hash comparisons. But in your case, the cost of creating a substring and calculating its hash (which involves some calculations for every character) is often greater than doing a direct string comparison (which can bail out at the first difference).
As the Wikipedia article that you linked to says, you'll want to use a rolling hash, a hashing algorithm that allows you to calculate the hash of the next substring with just a few operations, regardless of how long that substring is.
Also, as far as I can tell, storing string.Length in a local variable doesn't offer any performance improvements. It does make the code slightly less readable though, in my opinion.
Wouldn't it be a minor optimization if you check haystack[i] == needle[0] before you call Substring() and calculate the hash?:
if (haystack[i] == needle[0])
{
string tempStr = haystack.Substring(i, m);
var hashTemp = tempStr.GetHashCode();
if (hash == hashTemp)
{
return i;
}
}