[LeetCode]--44. Wildcard Matching

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).The function prototype should be:bool isMatch(const char *s, const char *p)Some examples:isMatch("aa","a") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false

Let’s see the answer!

public class Solution {    public boolean isMatch(String str, String pattern) {        int s = 0, p = 0, match = 0, starIdx = -1;        while (s < str.length()) {            // case that: str[s] == pattern[p] or pattern[p] == '?'            if (p < pattern.length() && (pattern.charAt(p) == '?' || str.charAt(s) == pattern.charAt(p))) {                s++;                p++;            }            // case that * found, only advance the pattern pointer.            else if (p < pattern.length() && pattern.charAt(p) == '*') {                starIdx = p;                match = s;                p++;            }            // case that pattern[p] != str[s]  &&  pattern[p] != '?'  &&   pattern[p] != '*', that             // means la            else if (starIdx != -1) {                p = starIdx + 1;                match++;                s = match;            }            // case that current pattern pointer is not *, and last pointer was not * either. So, don't match.            else return false;        }        // Let's check for remaining character in pattern        while (p < pattern.length() && pattern.charAt(p) == '*') {            p++;        }        return p == pattern.length();    }}

Dynamic Programing

Can also solvle it with DP method!
1. Python Code

class Solution:    # @param s, an input string    # @param p, a pattern string    # @return a boolean    def isMatch(self, s, p):        m,n = len(s),len(p)        cnt =p.count('*')        if n - cnt > m:            return False        dp=[True] + [False]*n        for j in range(1,n+1):            dp[j]= dp[j-1] and p[j-1]=='*'        for i in range(1,m+1):            cur=[False]*(n+1)            for j in range(1,n+1):                if p[j-1]=='*':                    cur[j]= cur[j-1] or dp[j]                elif p[j-1]==s[i-1] or p[j-1]=='?':                                       cur[j]=dp[j-1]               dp=cur        return dp[n]

1. Java Code

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

Reference

1. https://discuss.leetcode.com/topic/10794/my-java-dp-solution
2. https://discuss.leetcode.com/topic/3040/linear-runtime-and-constant-space-solution
3. https://www.hrwhisper.me/leetcode-regular-expression-matching-leetcode-wildcard-matching/