LeetCode刷题笔录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") → 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

递归,分为p[j] ='*'和!='*'两种情况。

1.p[j] != '*'

如果s[i] == p[j]（包括p[j]是？的情况），则递归下一层i+1和j+1

如果不相等则返回false

2.p[j] == '*'

则对于从i开始一直到s结尾的s的每一个substring都要进行匹配

public class Solution {
    public boolean isMatch(String s, String p) {
        return isMatch(s, p, 0, 0);
    }
    
    public boolean isMatch(String s, String p, int i, int j){
        if(j == p.length()){
            return i == s.length();
        }
        
        if(p.charAt(j) != '*'){
            if(i >= s.length() || (p.charAt(j) != s.charAt(i) && p.charAt(j) != '?'))
                return false;
            else
                return isMatch(s, p, i + 1, j + 1);
        }
        
        else{
            while(i < s.length()){
                if(isMatch(s, p, i, j + 1))
                    return true;
                i++;
            }
        }
        
        return isMatch(s, p, i, j + 1);
    }
    
    
}