LeetCode_44 Wildcard Matching

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本文深入探讨了如何通过优化回溯算法解决复杂正则表达式的匹配问题，特别关注了'?'和'*'字符的特殊用法。详细介绍了两种常见优化方法，并通过实例演示了这些方法在实际应用中的效果。此外，还提供了针对特定输入字符串的失败案例分析，以帮助读者更好地理解问题所在。

Link to original problem: 这里写链接内容

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”) → false
isMatch(“aa”,”aa”) → true
isMatch(“aaa”,”aa”) → false
isMatch(“aa”, “*”) → true
isMatch(“aa”, “a*”) → true
isMatch(“ab”, “?*”) → true
isMatch(“aab”, “c*a*b”) → false

Related problems:
10 Regular Expression Matching 这里写链接内容

I tried the basic backtracking (recursion) method and the method which has optimization on continuous ‘*’, for this problem and get TLE twice. So I came up with the idea dealing with this problem from the tail of those two string.

public class Solution {
    public boolean isMatch(String s, String p) {
        if(p.length() == 0) return (s.length() == 0);
        if(p.charAt(0) == '*'){
            if(s.length() == 0) return isMatch(s, p.substring(1));
            else return isMatch(s, p.substring(1)) || isMatch(s.substring(1), p);
        }
        else{
            if(s.length() == 0) return false;
            else if(s.charAt(0) == p.charAt(0) || p.charAt(0) == '?') return isMatch(s.substring(1), p.substring(1));
            else return false;
        }
    }
}

This code failed on:
“aaabbbaabaaaaababaabaaabbabbbbbbbbaabababbabbbaaaaba”
“a*******b”

public class Solution {
    public boolean isMatch(String s, String p) {
        if(p.length() == 0) return (s.length() == 0);
        if(p.charAt(0) == '*'){
            if(s.length() == 0) return isMatch(s, p.substring(1));
            else if(p.length() > 1 && p.charAt(1) == '*') return isMatch(s, p.substring(1));
            else return isMatch(s, p.substring(1)) || isMatch(s.substring(1), p);
        }
        else{
            if(s.length() == 0) return false;
            else if(s.charAt(0) == p.charAt(0) || p.charAt(0) == '?') return isMatch(s.substring(1), p.substring(1));
            else return false;
        }
    }
}

And this code failed on:
“babbbbaabababaabbababaababaabbaabababbaaababbababaaaaaabbabaaaabababbabbababbbaaaababbbabbbbbbbbbbaabbb”
“b**bb**a**bba*b**a*bbb**aba***babbb*aa****aabb*bbb***a”