LeetCode 44. Wildcard Matching

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本文介绍了一种实现通配符匹配的算法，该算法支持 '?' 和 '*' 两种通配符，其中 '?' 可以匹配任意单个字符，而 '*' 可以匹配任意长度的字符序列。文章提供了一个具体的 C++ 实现案例，并通过多个示例展示了该算法的工作原理。

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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") → 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

bool match(char a, char b) {
  return a == b || b == '?';
}

bool isMatch(string target, string pattern) {
  int m = target.size();
  int n = pattern.size();
  vector< vector<bool> > dp(m + 1, vector<bool>(n + 1));
  dp[0][0] = true;
  for(int i = 0; i <= m; ++i) {
    for(int j = 0; j <= n; ++j) {
      if(!i && !j) continue;
      if(i > 0 && j > 0) {
        if(match(target[i-1], pattern[j-1]) && dp[i-1][j-1]) // regular match
          dp[i][j]= true;
      }
      if(i > 0 && j > 0) {   //cover three, * matches 0(i, j - 1), * matches more than 1(i, j - 1), * matches 1, (i-1, j - 1)  
        if(pattern[j-1] == '*' && (dp[i-1][j] || dp[i][j-1] || dp[i-1][j-1])) dp[i][j] = true;
      }
      if(j > 0) { // cover one case. aaa, aaa* --> * matches 0.
          if(pattern[j-1] == '*' && dp[i][j-1]) dp[i][j] = true;
      }
    }
  }
  return dp[m][n];
}