Return to Revisions

The complexity of this method is actually \$O(n^2)\$, where \$n\$ is s1.Length (which is equal to s2.Length). Let's expand out IndexOf and see why.

foreach (char c in s1)
{
    int ix = -1;
    for (var i = 0; i < s2.Length; i++)
    {
        if (s2[i] == c)
        {
            ix = i;
            break;
        }
    }

    if (ix == -1)
        return false;
}

return true;

As @Pimgd pointed out, it is also incorrect.