10. Regular Expression Matching（很重要!!!）

10. Regular Expression Matching

Implement regular expression matching with support for '.' and '*'.

'.' Matches any single character.'*' Matches zero or more of the preceding element.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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

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This problem has a typical solution using Dynamic Programming. We define the state P[i][j] to be true if s[0..i) matches p[0..j)and false otherwise. Then the state equations are:

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 (p[j - 1] == '*'){dp[i][j] = dp[i][j - 2] ||(i > 0 && dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'));}else{if (i > 0 && (s[i-1] == p[j-1] || p[j-1] == '.')){dp[i][j] = dp[i - 1][j - 1];}}}}return dp[m][n];}};

44. Wildcard Matching

Implement wildcard pattern matching with support for '?' and '*'.

'?' 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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false

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这个跟上面的区别是，这里*是通配符可以匹配任何字符，而不是它前面的那个字符。

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 (p[j - 1] == '*'){dp[i][j] = dp[i][j - 1] || (i > 0 && dp[i - 1][j]);//}else{if (i > 0 && (s[i - 1] == p[j - 1] || p[j - 1] == '?')){dp[i][j] = dp[i - 1][j - 1];}}}}return dp[m][n];}};