10.正则表达式匹配

Regular Expression Matching

问题描述：

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

知识补充：

字符串操作:

s.substr(pos,len);//截取字符串,从下标为pos的位置截取（包括此位置）长度为len字符串
s.substr(pos);//从pos位置截取到最后
s1.equals(s2);//比较字符串 函数在>时返回1，<时返回-1，==时返回0
s1==s2;//比较字符串

逻辑符号运算顺序：

!>&&>||;//

测试代码：

本题适合使用动态规划办法求解，通过将问题划分为最优子结构，最终求得最优解。

分为如下三种情况,f[i][j]表示字符串s与p的匹配情况i，j分别为其字符串长度：

1 ) p[j-1]!=’*’:

f[i][j]=f[i-1][j-1]&&(s[i-1]==p[j-1]||p[j-1]==’.’)

2 ) p[j-1]==’*’:

A:*相当于匹配0次：

f[i][j]=f[i][j - 2]

B:*相当于重复匹配多次：

f[i][j]=s[i - 1] == p[j-2] && f[i - 1][j]

bool isMatch(string s, string p) {
if (p.empty())    return s.empty();
if(p[1]=='*')
{
return(isMatch(s, p.substr(2)) || !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p));

}
else if(p[0]=='.')
{
if(s.empty())
return false;
return(isMatch(s.substr(1),p.substr(1)));
}else {
if(s[0]==p[0])
{
return(isMatch(s.substr(1),p.substr(1)));
}else
{
return false;
}
}

}

性能：

参考答案：

if (p.empty())    return s.empty();

if ('*' == p[1])
// x* matches empty string or at least one character: x* -> xx*
// *s is to ensure s is non-empty
return (isMatch(s, p.substr(2)) || !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p));
else
return !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p.substr(1));
}

bool isMatch(string s, string p) {
int m = s.size(), n = p.size();
vector<vector<bool>> f(m + 1, vector<bool>(n + 1, false));

f[0][0] = true;
for (int i = 1; i <= m; i++)
f[i][0] = false;
// p[0.., j - 3, j - 2, j - 1] matches empty iff p[j - 1] is '*' and p[0..j - 3] matches empty
for (int j = 1; j <= n; j++)
f[0][j] = j > 1 && '*' == p[j - 1] && f[0][j - 2];

for (int i = 1; i <= m; i++)
for (int j = 1; j <= n; j++)
if (p[j - 1] != '*')
f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1] || '.' == p[j - 1]);
else
// p[0] cannot be '*' so no need to check "j > 1" here
f[i][j] = f[i][j - 2] || (s[i - 1] == p[j - 2] || '.' == p[j - 2]) && f[i - 1][j];

return f[m]
;

}