leetcode题解c++ | 10. Regular Expression Matching

题目：https://leetcode.com/problems/regular-expression-matching/#/description

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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

分析：

刚开始想直接匹配，但对于带*的，就很难做判断。想了很久才想通dp来做比较简单

f[i][j]: s[0..i-1] 是否匹配p[0..j-1]

if p[j - 1] != '*'

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

if p[j - 1] == '*', 设 p[j - 2] 为 x

      1) "x*" 重复0次：f[i][j] = f[i][j - 2]

      2)"x*" 重复1次以上：f[i][j] = f[i-1][j] && (p[j-2]=='.'||s[i-1]==p[j-2])

注意这里只需要转移重复一次的情况，而不需要枚举重复的次数。还有初始化需要分类，这个想一下也很容易想通。

c++实现：

bool isMatch(string s, string p){    int m=s.length(), n=p.length();    bool f[m+1][n+1];    f[0][0] = 1;    for(int i=1; i<=m; ++i)        f[i][0] = 0;    for(int i=1; i<=n; ++i)        if(p[i-1]!='*')            f[0][i] = 0;        else            f[0][i] = f[0][i-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]&& (p[j-1]=='.'||s[i-1]==p[j-1]);            else                f[i][j] = f[i][j-2] || (f[i-1][j]&& (p[j-2]=='.'||s[i-1]==p[j-2]));        }    return f[m][n];}