[LeetCode#44]Wildcard Matching

Problem:

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

Analysis:

Key:
For String problem, when the order of charcters in the string matters, we should try to take advantage of dynamic programming for future works. However, if it just care about (character set, could be permutated), we should try to take advantage of HashMap. 

Sadlly, Even though you can come up with the idea of using two dimensional dynamic method to solve this problem, you can still be stucked with a complex implementation routine.

For example, for this problem, you can come up with the following transitional equation:
1. if checkboard[i-1][j-1] == true, and s.charAt(i) == s.charAt(j) || p.charAt(j) == '?' || p.charAt(j) == '*', we have checkboard[i][j].
if (check_board[j] == true && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '?')
    flag = true;

2. if checkboard[i][j-1] == true, and s.charAt(j) == '*'
   if checkboard[i-1][j-1] == true, and s.charAt(j) == '*' (j could be used for matching " ")
   if checkboard[i-2][j-1] == true, ........................................................
 
According to my prgramming stereotype, it is very hard to implement 
if checkboard[i][j-1] == true, and s.charAt(j) == '*' (j could be used for matching " ")
if checkboard[i-1][j-1] == true, and s.charAt(j) == '*' (j could be used for matching " ")
if checkboard[i-2][j-1] == true, .....
KEY: Since '*' can match any sequence of characters. 

If you organize the loop through following way:
for (int i = 0; i < s.length(); i++) {
    for (int j = p.length() - 1; j >= 0; j++) {
        ...
    }
}
How do you retrieve the state of preivous rows in one dimensional array? Impossible!
Why not use j for outer lopp, and i for inner loop?
for (int j = 0; j < p.length(); j++) {
    for (int i = s.length() - 1; i >= 0; i--) {
    ...
    }
}


Skill:
a. properly use "" substring for both string. (This could lay a great foudation for our future inference)
boolean[] check_board = new boolean[p.length()+1];
check_board[0] = true; //cause "" = ""

b. understand the start and direction of inference.
 for (int j = 0; j < p.length(); j++) {
    for (int i = s.length() - 1; i >= 0; i--) {
    ...
    }
 }
Since we calculate the value for check_board[j][i], based on check_board[j-1][i],check_board[j-2][i] ...
We must do the caculation from "right to left". 
for (int i = s.length() - 1; i >= 0; i--)

c. update the corner(boundary) column of the check_board. 
true|flase|false|false|false
---------------------------
    |
    |
    |
    |
    |
c.1 the first row of the matrix, must only have the first element equal to true. 
c.2 the first column's value is pending. before the first non "*" charcter appears in the first column, all elements should be true. 
heck_board[0] = check_board[0] && (p.charAt(j) == '*');

d. use proper branch for organizing the program.
Since "*" is quite different from other characters appeared in the pattern string p, we not we distinguish and cope both cases seperately. How about distinguish "*" from other characters.
d.1 if p.charAt(j) != '*', 
check_board[i+1] = check_board[i] && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '?');   

d.1 if p.charAt(j) == '*',
as long as check_board[#][j-1] is true, all check_board[after #][j-1] shoud be true. Since '*' could equal to any sequence of characters. 
if (p.charAt(j) != '*') {
    ...
} else{
    int i = 0;
    while (i <= s.length() && !check_board[i]) {
        i++;
    }
    for (; i <= s.length(); i++) {
        check_board[i] = true;
    }               
}
Skil: we scan from check_board[i], but we acutally not change it!!! All elements are under same logic. How beautiful?



Errors: (my first solution):
public class Solution {
    public boolean isMatch(String s, String p) {
        if (s != null && p == null)
            return false;
        boolean[] check_board = new boolean[p.length()+1];
        check_board[0] = true;
        //be carefull with charAt when outside checking condition!
        if (p.length() > 0 && p.charAt(0) == '*') 
            check_board[1] = true;
        for (int i = 0; i < s.length(); i++) {
            for (int j = p.length() - 1; j >= 0; j++) {
                if (check_board[j] == true && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '?' || p.charAt(j) == '*'))
                    check_board[j+1] = true;
                if (check_board[j+1] == true && p.charAt(j) == '*')
                    check_board[j+1] = true;
            }
            check_board[0] = false; //wrong too!!!!
        }
        return check_board[p.length()];
    }
}

Input:
"aa", "a"
Output:
true
Expected:
false


Problemetic part:
for (int j = p.length() - 1; j >= 0; j++) {
    if (check_board[j] == true && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '?' || p.charAt(j) == '*'))
        check_board[j+1] = true;
    if (check_board[j+1] == true && p.charAt(j) == '*')
        check_board[j+1] = true;
}
Reason: what if check_board[i][j+1] has not been udpated for true?
if would contine to hold the same value as check_board[i-1][j+1].

Wrong solution: <When you use one-dimensional array to save space, must rember to udpate state!!!>
// The case is to hard to carry on any more!
Fix:
for (int j = p.length() - 1; j >= 0; j++) {
    boolean flag = false;
    if (check_board[j] == true && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '?' || p.charAt(j) == '*'))
        flag = true;
    if (check_board[j+1] == true && p.charAt(j) == '*')
        flag = true;
    check_board[j+1] = flag;
}

Solution:

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

 

转载于:https://www.cnblogs.com/airwindow/p/4759172.html