http://blog.csdn.net/hk2291976/article/details/51165010
dp[i][j]的含義是s[0-i] 與 s[0-j]是否匹配。
1 p.charAt(j) == s.charAt(i) : dp[i][j] = dp[i-1][j-1]
2 If p.charAt(j) == ‘.’ : dp[i][j] = dp[i-1][j-1];
3If p.charAt(j) == ‘’:
here are two sub conditions:
1 if p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2] //in this case, a only counts as empty
2 if p.charAt(i-1) == s.charAt(i) or p.charAt(i-1) == ‘.’:
dp[i][j] = dp[i-1][j] //in this case, a* counts as multiple a
dp[i][j] = dp[i][j-1] // in this case, a* counts as single a
dp[i][j] = dp[i][j-2] // in this case, a* counts as empty
class Solution
{
public:
static const int FRONT=-1;
bool isMatch(string s, string p)
{
int m = s.length(),n = p.length();
bool dp[m+1][n+1];
dp[0][0] = true;
//初始化第0行,除了[0][0]全為false，毋庸置疑，因為空串p只能匹配空串，其他都無能匹配
for (int i = 1; i <= m; i++)
dp[i][0] = false;
//初始化第0列，只有X*能匹配空串，如果有*，它的真值一定和p[0][j-2]的相同（略過它之前的符號）
for (int j = 1; j <= n; j++)
dp[0][j] = j > 1 && '*' == p[j - 1] && dp[0][j - 2];
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
if (p[j - 1] == '*')
{
dp[i][j] = dp[i][j - 2] || (s[i - 1] == p[j - 2] || p[j - 2] == '.') && dp[i - 1][j];
}
else //只有當前字符完全匹配，才有資格傳遞dp[i-1][j-1] 真值
{
dp[i][j] = (p[j - 1] == '.' || s[i - 1] == p[j - 1]) && dp[i - 1][j - 1];
}
}
}
return dp[m][n];
}
};
