Regular Expression Matching

Description

Implement regular expression matching with support for '.' 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", "ca*b") → true

Hint

Train of Thought

This Solution use 2D DP. beat 90% solutions, very simple.

Here are some conditions to figure out, then the logic can be very straightforward.

1, If p.charAt(j) == s.charAt(i) : dp[i][j] = dp[i-1][j-1]; 2, If p.charAt(j) == '.' : dp[i][j] = dp[i-1][j-1]; 3, If p.charAt(j) == '': here are two sub conditions: 1 if p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2] //in this case, a only counts as empty 2 if p.charAt(i-1) == s.charAt(i) or p.charAt(i-1) == '.': dp[i][j] = dp[i-1][j] //in this case, a counts as multiple a or dp[i][j] = dp[i][j-1] // in this case, a counts as single a or dp[i][j] = dp[i][j-2] // in this case, a* counts as empty

Code

 public boolean isMatch(String s, String p) {

if (s == null || p == null) {
    return false;
}
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]) {
        dp[0][i+1] = true;
    }
}
for (int i = 0 ; i < s.length(); i++) {
    for (int j = 0; j < p.length(); j++) {
        if (p.charAt(j) == '.') {
            dp[i+1][j+1] = dp[i][j];
        }
        if (p.charAt(j) == s.charAt(i)) {
            dp[i+1][j+1] = dp[i][j];
        }
        if (p.charAt(j) == '*') {
            if (p.charAt(j-1) != s.charAt(i) && p.charAt(j-1) != '.') {
                dp[i+1][j+1] = dp[i+1][j-1];
            } else {
                dp[i+1][j+1] = (dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1]);
            }
        }
    }
}
return dp[s.length()][p.length()];

}

Complexity