Leetcode: 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") → 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)

?

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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()];     } }