[LeetCode]Wildcard Matching

class Solution {//1. No match. Simply return false. This is the last case in the program.//2. Single match. Either *s == *p, or *p == '?'. The return value of this case depends on the //result of the rest parts of both s and p (recursion call), which start at the 'diagonal' position by advancing both s and p by 1 (++s, ++p)//3. Star case, i.e. when *p == '*'. This is where the complication comes in. //A star can match 0, 1, 2, ..., to the end of string s. So, as long as one of the substrings match (recursion call), //after advance over '*', it returns true. This case returns false only after exhausting all possible substrings without a match.//update row by row, should practice more, DP, the time complexity is O(n*m)public:bool isMatch(const char *s, const char *p) {// Start typing your C/C++ solution below// DO NOT write int main() functionif (!*s && !*p) return true;int ms_max = 1;//size of *sconst char* ss = s;while(*ss){ ++ms_max; ++ss;}int np_max = 1;const char* pp = p;while(*pp){ if(*pp!='*') ++np_max; ++pp;}if(ms_max < np_max) return false;vector<vector<bool> > r(2, vector<bool>(ms_max, false));//for the sake of saving space bool flag = 1;r[0][0] = true;do{//*pint ms = 1;ss = s;if (*p == '*'){while(*p == '*') ++p;//skip consecutive star--p;r[flag][0] = r[!flag][0];//from the previous row, because star can contain zero characterfor( ;ms <= ms_max; ++ms){//up and leftif (r[!flag][ms] || r[flag][ms-1]) r[flag][ms] = true;else r[flag][ms] = false;}}else{do{//for every character in p, update the row bool r_flag = false;if (*ss == *p || *p == '?')r_flag = r[!flag][ms-1];//diagonalr[flag][ms]=r_flag;++ms;++ss;}while(*ss);//*sr[flag][0] = false;//there must be some character in ss, if current character in p is not star}++p;flag = !flag;//very fantastic}while(*p);return r[!flag][ms_max-1];}};

second time

class Solution {//if strlen is used, then it will be TLE//iteration based solutionpublic:    bool isMatch(const char *s, const char *p) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        //int len1 = strlen(s);        //int len2 = strlen(p);                bool star = false;        const char* starPtr = NULL;        const char* savePtr = NULL;        while(*s != '\0')        {            if(*p == '?') s++, p++;            else if(*p == '*')            {                while(*p == '*') p++;                p--;                starPtr = p;                p++;                savePtr = s;                star = true;            }            else             {                if(*s == *p) s++, p++;                else                {                    if(star == true) s = ++savePtr, p = starPtr+1;//depend on star                    else return false;                }            }        }        while(*p == '*') p++;        return (*s == '\0' && *p == '\0');    }};