44. Wildcard Matching

• Difficulty: Hard

• Topics: String, Dynamic Programming, Backtracking, Greedy

• Similar Questions:

  • Regular Expression Matching

Problem:

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

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 = "*"
Output: true
Explanation: '*' matches any sequence.

Example 3:

Input:
s = "cb"
p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.

Example 4:

Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".

Example 5:

Input:
s = "acdcb"
p = "a*c?b"
Output: false

Solutions:

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.length();
        int 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 = 0; j <= n; ++j) {
                if (i == 0 && j == 0)   continue;
                if (j == 0) continue;

                if (p[j - 1] == '*') {
                    dp[i][j] = dp[i][j-1];
                    if (i > 0) {
                        dp[i][j] = dp[i][j] || dp[i-1][j];
                    }
                } else if (i > 0 && (p[j - 1] == '?' || s[i-1] == p[j-1])) {
                    dp[i][j] = dp[i-1][j-1];
                }
            }
        }
        return dp[m][n];
    }
};

More concise DP

It is not necessary to separate the initialization process.

From [Grandyang] (https://www.cnblogs.com/grandyang/p/4461713.html)

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.size(), n = p.size();
        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[m][n];
    }
};