LeetCode 44 Wildcard Matching

Brute Force:                                                                                               has '*' before -----   find last index of s and p when star found, s move forward

                                  char s  not reaches end  -- >  p reaches end   

                                                |                                                                  no  star -----  NOT MATCH
                                                |   same?

                                 no       /       \    yes

                        is p '?"        |               s++ p++

                           no       /         \   yes

                      is p '*'      |             s++, p++

                     no       /          \   yes

has '*' before          |             skip all p star and record the index of s and p

              no         /            \  yes

          NOT MATCH         find last index of s and p when star found

                                              s move forward

 

class Solution {
    public boolean isMatch(String s, String p) {
        int prevS = 0;
        int prevP = 0;
        
        int currentS = 0;
        int currentP = 0;
        boolean star = false;
        while (currentS < s.length()) {
            if (currentP >= p.length()) {
                if (star) {
                    currentS = ++prevS;
                    currentP = prevP;
                } else {
                    return false;
                }            }
            if (s.charAt(currentS) != p.charAt(currentP)) {
                if (p.charAt(currentP) == '?') {
                    currentS++;
                    currentP++;
                } else if (p.charAt(currentP) == '*') {
                    while (currentP < p.length() && p.charAt(currentP) == '*') {
                        currentP++;
                    }
                    if (currentP == p.length()) {
                        return true;
                    }
                    star = true;
                    prevS = currentS;
                    prevP = currentP;
                } else if (star) {
                    currentS = ++prevS;
                    currentP = prevP;
                } else {
                    return false;
                }
            } else {
                currentS++;
                currentP++;
            }
        }
        
        while (currentP < p.length() && p.charAt(currentP) == '*') {
            currentP++;
        }
        return currentP == p.length();
    }
}

 

 

DP:

       For dp[i][j] = dp[i + 1][j] || dp[i][j + 1] means either there is a point k that s(k , end) matches p(j + 1, end) when k > i or * matches 0 char here. So s(i , end) matches p(j + 1, end)

 

class Solution {
    public boolean isMatch(String s, String p) {
        boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
        dp[s.length()][p.length()] = true;
        for (int i = p.length() - 1; i >= 0; i--) {
            if (p.charAt(i) != '*') {
                break;
            }
            dp[s.length()][i] = true;
        }
        
        for (int i = s.length() - 1; i >= 0; i--) {
            for (int j = p.length() - 1; j >= 0; j--) {
                if (p.charAt(j) == '?' || p.charAt(j) == s.charAt(i)) {
                    dp[i][j] = dp[i + 1][j + 1];
                } else if (p.charAt(j) == '*') {
                    dp[i][j] = dp[i + 1][j] || dp[i][j + 1];
                } else {
                    dp[i][j] = false;
                }
                
            }
        }
        return dp[0][0];
    }
}