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[Leetcode] Regular Expression Matching

2015-01-28 13:44 288 查看

Implement regular expression matching with support for

'.'

and

'*'

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

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", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

注意：这里的a*表示a可以重复0次或者多次，不是a和*分开的。

It seems that some readers are confused about why the regex pattern

".*"

matches the string

"ab"

".*"

means repeat the preceding element 0 or more times. Here, the "preceding" element is the dot character in the pattern, which can match any characters. Therefore, the regex pattern

".*"

allows the dot to be repeated any number of times, which matches any string (even an empty string). Think carefully how you would do matching of

'*'

.Please note that
'*'
in regular expression is different from wildcard matching, as we match the previous character 0 or more times. But, how many times? If you are stuck,recursion is your friend.

动态规划，无敌存在

class Solution {
public:
bool isMatch(const char *s, const char *p) {
int i, j;
int m = strlen(s);
int n = strlen(p);

/**
* b[i + 1][j + 1]: if s[0..i] matches p[0..j]
* if p[j] != '*'
* b[i + 1][j + 1] = b[i][j] && s[i] == p[j]
* if p[j] == '*', denote p[j - 1] with x,
* then b[i + 1][j + 1] is true iff any of the following is true
* 1) "x*" repeats 0 time and matches empty: b[i + 1][j -1]
* 2) "x*" repeats 1 time and matches x: b[i + 1][j]
* 3) "x*" repeats >= 2 times and matches "x*x": s[i] == x && b[i][j + 1]
* '.' matches any single character
*/
bool b[m + 1][n + 1];
b[0][0] = true;
for (i = 0; i < m; i++) {
b[i + 1][0] = false;
}
// p[0..j - 2, j - 1, j] matches empty iff p[j] is '*' and p[0..j - 2] matches empty
for (j = 0; j < n; j++) {
b[0][j + 1] = j > 0 && '*' == p[j] && b[0][j - 1];
}

for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
if (p[j] != '*') {
b[i + 1][j + 1] = b[i][j] && ('.' == p[j] || s[i] == p[j]);
} else {
b[i + 1][j + 1] = b[i + 1][j - 1] && j > 0 || b[i + 1][j] ||
b[i][j + 1] && j > 0 && ('.' == p[j - 1] || s[i] == p[j - 1]);
}
}
}
return b[m]
;
}
};

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