10. Regular Expression Matching

​​'.'​​ 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") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true


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​​P[i][j]​​ to be ​​true​​ if ​​s[0..i)​​ matches ​​p[0..j)​​and ​​false​​​​P[i][j] = P[i - 1][j - 1]​​

  1. , if 

​​p[j - 1] != '*' && (s[i - 1] == p[j - 1] || p[j - 1] == '.')​​

  1. ;

​​P[i][j] = P[i][j - 2]​​

  1. , if 

​​p[j - 1] == '*'​​

  1.  and the pattern repeats for 

​​0​​​​P[i][j] = P[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.')​​

  1. , if 

​​p[j - 1] == '*'​​

  1.  and the pattern repeats for at least 

​​1​​

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



​​'?'​​ 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") → 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



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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];
  }
};