Leet Code Regular Expression Matching Problem - c++

Question

https://leetcode.com/problems/regular-expression-matching
I was doing this practice problem (cpp) and while faster solutions are in the comments, I would like to understand why my code isn't working. This fails with s = "mississippi" and p = "mis*is*p*.". Tracing through the code, I figured it would correctly remove the first two letters, then when seeing the s* it would go through the s in the string (two of them), then remove the i in both, remove all the s (again 2) then remove all the p's (which is none, because it's compared against the i in the first string, so it should not modify that string). Finally, the '.' would match with the first p and remove both. So the final string should be "pi" and return false when the length is compared to zero.
class Solution {
public:
bool isMatch(string s, string p) {
while (s.length() > 0){
if (p.length() == 0){
return false;
}else if (p.length() == 1){
return p.compare(s) == 0 || p.at(0) == '.';
}else{
if (p.at(1) == '*'){
char c = p.at(0);
p = p.substr(2);
if (c == '.'){
return true;
}
int spot = 0;
while(spot < s.length() && s.at(spot) == c){
spot++;
}
if (spot != 0){
s = s.substr(spot);
}
}else{
if (s.at(0) != p.at(0) && p.at(0) != '.'){
return false;
}
s = s.substr(1);
p = p.substr(1);
}
}
}
return s.length() == 0;
}
};

Answer 1

Your logic is faulty here
return p.compare(s) == 0 || p.at(0) == '.';
That should be
return p.compare(s) == 0 || (s.length() == 1 && p.at(0) == '.');
That took me five minutes to find, two minutes looking at the code without seeing the problem, and then three minutes using a debugger to track down the logic error. You really should learn to use a debugger, much more efficient than asking on SO.
Some tips here.