跳过Java if语句

Question

The method below takes in a string and a pattern and returns true if they match each other. A '.' matches 1 char and a '*' matches 0 or more (eg expMatch("abc", "ac") should return true ). I added a bunch of print statements to see where I went wrong and it seems like the if statement is being skipped even if the str.length() == 1 .

I call it with System.out.println(expMatch("abc", "a*c"));

Here is the code:

public static boolean expMatch(String str, String pat)
{   
    if (str.charAt(0) == pat.charAt(0) || pat.charAt(0) == '.')
    {
        System.out.println("in if");
        System.out.println(str.charAt(0));
        System.out.println(pat.charAt(0));
        System.out.println(str.length());
        if (str.length() == 1)
            return true;
        expMatch(str.substring(1), pat.substring(1));
    }

    else if (pat.charAt(0) == '*')
    {   
        System.out.println("in else");
        System.out.println(str.charAt(0));
        System.out.println(pat.charAt(0));
        if (str.length() == 1)
            return true;
        if (str.charAt(0) == pat.charAt(1)) //val of * = 0
            expMatch(str, pat.substring(1));
        else if (str.charAt(1) ==pat.charAt(1))
            expMatch(str.substring(1), pat.substring(1));           
    }

    return false;           
}

and the output is:

in if
a
a
3
in else
b
*
in if
c
c
1
false

Even if the length is 1 it skips the if? Any idea why? PS I'm not looking for the solution, just why the if statement is being skipped.

Answer 1

You always return false from the method at the very end. You are calling expmatch recursively but never using the return value. The code comes in to the first if, recurses (because length is not 1) and upon returning will go to the final return statement which returns false.

Answer 2

You need to add a return before your expMatch() calls - because the false comes from your last line return false;

What happens is this:

• you call expMatch() with the two Strings.
• you enter the if clause
• the if clause enters expMatch() recursively
• you enter the else clause
• the else clause enters expMatch() recursively
• you enter the if clause again
• you leave the expMatch() method
• you leave the other expMatch method
• false is returned

Answer 3

Your approach is logically incorrect even if you apply the fixes the others suggested. Try this test case:

System.out.println(expMatch("abddddc", "a*c"));

This is because when you encounter a * in the pattern, you have no way to know how many characters "to eat" from the search string.

To say the least, you need a loop somewhere, not just an if . Let me try to fix it for you (not sure if it's possible though, not sure if you always know which path to take, I mean in your recursion). Think some more about it. Here is another unpleasant test case:

System.out.println(expMatch("adddcac", "a*c")); 
// the * needs to eat dddca (despite the c present in dddca),  
// it should not stop recursing there at that c  

I think you need some sort of full search here.
Just an if or a while loop is not good enough.

EDIT : Here is a fixed version with a bunch of nasty tests. I think this is called non-linear recursion (as it's not a single path you try). Not 100% sure though about that term.

public class Test055 {

    public static void main(String[] args) {
        // System.out.println(expMatch("abddddc", "a*c"));

        System.out.println(expMatch("adcax", "a*c"));
        System.out.println(expMatch("adcax", "a*c*"));

        System.out.println(expMatch("adcacm", "*"));
        System.out.println(expMatch("adcacmmm", "a*c"));
        System.out.println(expMatch("adcacmmmc", "a*c"));
        System.out.println(expMatch("adcac", "a*c"));

        System.out.println(expMatch("adcacxb", "a*c.b"));
        System.out.println(expMatch("adcacyyb", "a*c.b"));
        System.out.println(expMatch("adcacyyb", "a*c*b"));

    }

    public static boolean expMatch(String str, String pat)
    {
        // System.out.println("=====================");
        // System.out.println("str=" + str);
        // System.out.println("pat=" + pat);

        if (pat.length() == 0 && str.length() > 0) {
            return false;
        } else if (pat.length() == 0 && str.length() == 0) {
            return true;
        } else if (pat.charAt(0) == '.'){
            return str.length() >= 1 && expMatch(str.substring(1), pat.substring(1));
        }else if (pat.charAt(0) != '*'){
            return str.length() >= 1 && pat.charAt(0) == str.charAt(0) && expMatch(str.substring(1), pat.substring(1));
        }else{
            // Now let's handle the tricky part

            // (1) Look for the 1st non-star in pattern
            int k=-1;
            char ch = ' ';
            for (int i=0; i<pat.length(); i++){
                if (pat.charAt(i) != '*'){
                    k = i;
                    ch = pat.charAt(k);
                    break;
                }
            }

            if (k==-1){
                // (2A) only stars found in pattern, OK, any str matches that
                return true;
            }else{
                // (2B) do full search now checking all  
                // possible candidate chars in str that 
                // match the char ch from pattern
                for (int i=0; i<str.length(); i++){
                    if (str.charAt(i)==ch){
                        boolean b = expMatch(str.substring(i+1), pat.substring(k+1));
                        if (b) return true;
                    }
                }
                return false;
            }
        }
    }
}