10. Regular Expression Matching

Given an input string (s) and a pattern (p), 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).

Note:

• s could be empty and contains only lowercase letters a-z.

• p could be empty and contains only lowercase letters a-z, and characters like . or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

第一個是利用DP的方法，定義一個2D的dp table，dp[i][j]代表s[0...i - 1]和p[0...j - 1]是否match。下方為動態規劃的關係式： 1. dp[i][j] = dp[i - 1][j - 1] if p[j - 1] != '*' and (s[i - 1] == p[j - 1] || p[j - 1] == '.') 2. dp[i][j] = dp[i][j - 2] if p[j - 1] == '*' and the pattern repeats for 0 times 3. dp[i][j] = dp[i - 1][j] if p[j - 1] == '*' and the pattern repeats for at least one times

// Dynamic programming
bool isMatch(string s, string p) { // time: O(m * n); space: O(m * n)
    int m = s.length(), n = p.length();
    vector<vector<bool> > dp(m + 1, vector<bool>(n + 1, false));
    dp[0][0] = true;
    for (int i = 0; i <= m; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (j > 1 && p[j - 1] == '*') {
                dp[i][j] = dp[i][j - 2] || (i > 0 && (s[i - 1] == p[j - 2] || p[j - 2] == '.') && dp[i - 1][j]);
            } else {
                dp[i][j] = i > 0 && dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
            }
        }
    }
    return dp.back().back();
}

把i = 0的row先拿出來initialize，這樣後面的code就可以從i = 1開始loop，就不需要再確認i > 0。

// Dynamic programming
bool isMatch(string s, string p) { // time: O(m * n); space: O(m * n)
    int m = s.length(), n = p.length();
    vector<vector<bool> > dp(m + 1, vector<bool>(n + 1, false));
    dp[0][0] = true;
    for (int j = 1; j <= n; ++j) 
        dp[0][j] = j > 1 && dp[0][j - 2] && p[j - 1] == '*';
    for (int i = 1; i <= m; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (p[j - 1] == '*') {
                dp[i][j] = dp[i][j - 2] || (dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'));
            } else {
                dp[i][j] = dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
            }
        }
    }
    return dp.back().back();
}

// Recursion
bool isMatch(string s, string p) { // time: O(m * n); space: O(min(m, n))
    if (p.empty()) return s.empty();
    if (p.length() > 1 && p[1] == '*') {
        return isMatch(s, p.substr(2)) || 
            (!s.empty() && (s[0] == p[0] || p[0] == '.') && isMatch(s.substr(1), p));
    } else {
        return !s.empty() && (s[0] == p[0] || p[0] == '.') && isMatch(s.substr(1), p.substr(1));
    }
}