44. Wildcard Matching

'?' 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

we use dp to solve this problem;

let dp[i][j] be the boolean that means whether S(0 -> i) and P(0 -> j) is a match:

if(P[j] == '?') dp[i][j] = dp[i - 1][j - 1];

if(P[j] == '*') dp[i][j] = dp[i - 1][j] || dp[i][j - 1];

else dp[i][j] = (S[i] == P[j]) && dp[i - 1][j - 1];

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

dp[i][0] = false;

dp[0][j] 如果j之前都是＊就为true

code:

public class Solution {
public boolean isMatch(String s, String p) {
boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
dp[0][0] = true;
for(int i = 0; i < p.length(); i++){
if(p.charAt(i) == '*') dp[0][i + 1] = true;
else break;
}

for(int i = 1; i<=s.length(); i++){
for(int j = 1; j<=p.length(); j++){
char c = p.charAt(j - 1);
if(c == '?'){
dp[i][j] = dp[i - 1][j - 1];
} else if (c == '*'){
dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
} else {
dp[i][j] = (s.charAt(i - 1) == c) && dp[i - 1][j - 1];
}
}
}

return dp[s.length()][p.length()];
}
}