44. Wildcard Matching *HARD*

'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).

The matching should cover the entire input string (not partial).

The function prototype should be:
bool isMatch(const char *s, const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false

1. 动态规划

bool isMatch(string s, string p) {
int ls = s.length(), lp = p.length(), i, j;
vector<vector<bool>> dp(2, vector<bool>(lp+1, 0));
bool k = 1;
dp[0][0] = 1;
for(i = 1; i <= lp; i++)
dp[0][i] = dp[0][i-1] && '*' == p[i-1];
for(i = 1; i <= ls; i++)
{
dp[k][0] = 0;
for(j = 1; j <= lp; j++)
{
if('*' == p[j-1])
dp[k][j] = dp[k][j-1] || dp[!k][j];
else
dp[k][j] = dp[!k][j-1] && (p[j-1] == s[i-1] || '?' == p[j-1]);
}
k = !k;
}
return dp[!k][lp];
}

bool isMatch(string s, string p) {
int ls = s.length(), lp = p.length(), last_i = -1, last_j = -1, i = 0, j = 0;
while(s[i])
{
if('*' == p[j])
{
j++;
if(!p[j])
return 1;
last_i = i;
last_j = j;
}
else if(s[i] == p[j] || '?' == p[j])
{
i++;
j++;
}
else if(last_i != -1)
{
i = ++last_i;
j = last_j;
}
else
return 0;
}
while('*' == p[j])
j++;
return !p[j];
}