10.Regular Expression Matching

定义一个二维的DP数组,其中dp[i][j]表示s[0,i)和p[0,j)是否match,然后有下面三种情况
1.  P[i][j] = P[i - 1][j - 1], if p[j - 1] != '*' && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
2.  P[i][j] = P[i][j - 2], if p[j - 1] == '*' and the pattern repeats for 0 times;
3.  P[i][j] = P[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'), if p[j - 1] == '*' and the pattern repeats for at least 1 times.
class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.size(), n = p.size();
        vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false));
        dp[0][0] = true;
        for (int i = 0; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (j > 1 && p[j - 1] == '*') {
                    dp[i][j] = dp[i][j - 2] || (i > 0 && (s[i - 1] == p[j - 2] || p[j - 2] == '.') && dp[i - 1][j]);
                } else {
                    dp[i][j] = i > 0 && dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
                }
            }
        }
        return dp[m][n];
    }
};
posted @ 2019-04-08 17:22  JohnRed  阅读(76)  评论(0编辑  收藏  举报