Method 2 LeetCode solving dynamic programming problems [Reprinted from LeetCode]

1. Subject No.

10- regular expression matching (difficult)

2. Method Description

It uses dynamic programming to determine the transfer equation dp [i] [j] =? .

3.10 topics and ideas and code that question

topic:

Give you a character string s and a law p, invite you to implement a support '' and '*' in the regular expression matching.

'.' Matches any single character
'*' matches zero or more of the preceding element that a
so-called matching, to cover the entire string s, and not part of the string.

Description:

s may be empty, and only lowercase letters az from the.
p may be empty and contain only lowercase letters from az, and characters. and *.
Example 1:

Input:
S = "AA"
P = "A"
Output: false
interpretation: "a" can not match the entire string "aa".
Example 2:

Input:
S = "AA"
P = "A *"
Output: true
explanation: because the '*' matches zero or representatives of the foregoing that a plurality of elements, this is in front of the element 'a'. Thus, the string "aa" may be regarded as 'a' is repeated once.
Example 3:

Input:
S = "ab &"
P =. " "
Output: true
interpretation:. " " Represents zero or more matches ( '*') of any character ( '.').
Example 4:

Input:
S = "AAB"
P = "C A B"
Output: true
explanation: because the '*' means zero or more, where 'c' is 0, 'a' is repeated once. So it can match the string "aab".
Example 5:

Input:
S = "Mississippi"
P = "MIS IS P *."
Output: false

Transfer equation solving ideas

Defined dp [i] [j] denotes the front p s i and the first j characters of characters matches
Case 1 , if s [i]p [j], or p [j]'.' (Since '' matches any character, it can be considered equal)

String \ Location	0	…	i
s	…	…	a

String \ Location	0	…	j
p	…	…	.

In this case dp [i] [j] = dp [i - 1] [j - 1]

Case 2 , when the p [j] == '*', the following will happen:
!! '.' When 2-1 s [i] = p [j - - 1], and p [j 1] = (* preceding character string does not match the character s)

String \ Location	0	…	i
s	…	…	a

String \ Location	0	…	j - 1	j
p	…	…	b	*

Because the '*' matches zero or representatives of the foregoing that a plurality of elements, where only represent 0

In this case dp [i] [j] = dp [i] [j - 2]

2-2 This s [i]p [j - 1], or p [j - 1]'.' ,The following figure:

String \ Location	0	…	i
s	…	…	a

String \ Location	0	…	j - 1	j
p	…	…	a	*

Here divided several situations:
2-2-1 when the following pictures, s [i] and p - string matching before [j 2], * Match 0

String \ Location	0	…	i - 1	i
s	…	…	a	b

String \ Location	0	…	j-3	j - 2	j-1	j
p	…	…	a	b	b	*

When the image below 2-2-2, s [i - 1] and p - string matching before [j 2], * Match 1

String \ Location	0	…	i - 1	i
s	…	…	a	b

String \ Location	0	…	j - 2	j-1	j
p	…	…	a	b	*

When the image below 2-2-3, s [i - 1] and when p [j] in the previous string matching, matching times *

String \ Location	0	…	i-2	i - 1	i
s	…	…	a	b	b

String \ Location	0	…	j - 2	j-1	j
p	…	…	a	b	*

Or the three cases of the relationship, the relationship is as follows:
DP [I] [J] DP = [I -. 1] [J] || DP [I -. 1] [J - 2] || DP [I] [ j - 2];
boundary value initialization
1. before performing the above calculation, all values of the array needs to be initialized to false,
2.dp [0] [0] = to true; it represents front before and s 0 p 0 characters a certain character match
3. when calculating i == dp value of 0, a case is also possible to match all match 0 * to:

String \ Location
s

String \ Location	0	1	2	3	4	5
p	a	*	b	*	c	*

Code

bool isMatch(char * s, char * p){
    int i, j, len1, len2;
    len1 = strlen(s);
    len2 = strlen(p);
    bool** dp;
    dp = (bool**)malloc(sizeof(bool*) * (len1 + 1));
    for (i = 0; i < len1 + 1; i++) {
        dp[i] = (bool*)malloc(sizeof(bool) * (len2 + 1));
        memset(dp[i], 0, sizeof(bool) * (len2 + 1));
    }
    dp[0][0] = true;
    for (i = 0; i < len2; i++) {
        if (p[i] == '*' && i > 0 && dp[0][i - 1]) {
            dp[0][i + 1] = true;
        }
    }
    for (i = 0; i < len1; i++) {
        for (j = 0; j < len2; j++) {
            if (s[i] == p[j] || p[j] == '.') {
                dp[i + 1][j + 1] = dp[i][j];
                continue;
            }
            if (p[j] == '*' && j > 0) {
                if (s[i] == p[j - 1] || p[j - 1] == '.') {
                    dp[i + 1][j + 1] = dp[i][j + 1] || dp[i][j - 1] || dp[i + 1][j - 1];
                } else {
                    dp[i + 1][j + 1] = dp[i + 1][j - 1];
                }
            }
        }
    }
    return dp[len1][len2];
}

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/regular-expression-matching
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