[LeetCode-Algorithms-10] "Regular Expression Matching" (2017.10.12-WEEK6)

题目链接：Regular Expression Matching

题目描述：

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

(1)思路：参考题后讨论：This problem has a typical solution using Dynamic Programming. We define the state P[i][j] to be true if s[0..i) matches p[0..j) and false otherwise. Then the state equations are:

1.P[i][j] = P[i - 1][j - 1], if p[j - 1] != ‘*’ && (s[i - 1] == p[j - 1] || p[j - 1] == ‘.’);

2.P[i][j] = P[i][j - 2], if p[j - 1] == ‘*’ and the pattern repeats for 0 times;

3.P[i][j] = P[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == ‘.’), if p[j - 1] == ‘*’ and the pattern repeats for at least 1 times.

维护一个二维bool数组（vector容器）。

(2)代码：

class Solution {
public:
bool isMatch(string s, string p) {
int m = s.length(), n = p.length();
vector<vector<bool> > dp(m + 1, vector<bool> (n + 1, false));
dp[0][0] = true;
for (int i = 0; i <= m; i++)
for (int j = 1; j <= n; j++)
if (p[j - 1] == '*')
dp[i][j] = dp[i][j - 2] || (i > 0 && (s[i - 1] == p[j - 2] || p[j - 2] == '.') && dp[i - 1][j]);
else dp[i][j] = i > 0 && dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
return dp[m]
;
}
};

(3)提交结果：