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
```

**Understand the problem:**

The key of the problem is to understand the '*', which is able to match ANY sequence of characters. e.g. isMatch(abcd, *) -> true. Note that * does not require the same character in the sequence, as was required by the regular expression matching.

We use DP to solve this problem.

-- Define dp[s.length() + 1][p.length() + 1], where dp[i[j] means the first i characters in string s matches the first characters of string p.

-- Initialization: dp[0][0] = true;

dp[i][0] = false;

dp[0][j] = dp [0][j - 1] IFF p.charAt(j - 1) == '*'

-- Transit function:

-- If p.charAt(j - 1) != '*', then dp[i][j] = dp[i - 1][j - 1] IFF s.charAt(i) == p.charAt(j) || p.charAt(j) == '?'

-- If p.charAt(j - 1) == '*', then

-- dp[i][j] = dp[i - 1][j - 1] || // Match 1 character

= dp[i][j - 1] || // Match 0 character

= dp[i - 1][j] // Match any sequence of characters

-- Return dp[s.length()][p.length()].

**Code (Java)**

public class Solution { public boolean isMatch(String s, String p) { if (p == null || p.length() == 0) { return s == null || s.length() == 0; } int rows = s.length(); int cols = p.length(); boolean[][] dp = new boolean[rows + 1][cols + 1]; dp[0][0] = true; for (int j = 1; j <= cols; j++) { if (p.charAt(j - 1) == '*') { dp[0][j] = dp[0][j - 1]; } } for (int i = 1; i <= rows; i++) { for (int j = 1; j <= cols; j++) { if (p.charAt(j - 1) != '*') { if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '?') { dp[i][j] = dp[i - 1][j - 1]; } } else { dp[i][j] = dp[i - 1][j - 1] || dp[i][j - 1] || dp[i - 1][j]; } } } return dp[s.length()][p.length()]; } }

why is dp[0][j] = dp [0][j - 1] IFF p.charAt(j - 1) == '*'?

ReplyDeletewhy is dp[0][j] = dp [0][j - 1] IFF p.charAt(j - 1) == '*'?

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