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
问题分析： 参考串的元素只能是英文字母，输入串的元素既能是字母也可以是”.”和“”，其中“.”能和任何字母进行匹配，也就是通配符；但是输入串不能以“ ”开头，因为“*”的用途是重复其前面相邻字母0~n次，因此前面必须存在其他字符. 思路分析：采用DP能解此题 记f[i][j]为s[0~i-1]与p[0~i-1]是否匹配的标识，当s[0~i-1]=p[0~i-1]时，f[i][j]=1,否则f[i][j]=0 现在来考虑f[i][j]的状态转移方程 (1)当p[j-1]!=*时，s[j-1]=p[j-1]且f[i-1][j-1]=1时，f[i][j]=1,否则f[i][j]=0; (2)当p[j-1]=*时，又分为两种情形 A)p[j-2]重复了零次，此时s[0~i-1]与p[0~i-1]匹配成功与否取决于s[0~i-1]=？p[0~i-3]，即f[i][j]=f[i][j-2]; B)p[j-2]至少被重复了一次，此时匹配成功的一个条件是s[i-1]=p[j-2]，另外f[i-1][j]必须为1，即f[i][j]=s[i-1]=p[j-2] &&f[i-1][j];
class Solution { public: bool isMatch(string s, string p) { /** * f[i][j]: if s[0..i-1] matches p[0..j-1] * if p[j - 1] != '*' * f[i][j] = f[i - 1][j - 1] && s[i - 1] == p[j - 1] * if p[j - 1] == '*', denote p[j - 2] with x * f[i][j] is true iff any of the following is true * 1) "x*" repeats 0 time and matches empty: f[i][j - 2] * 2) "x*" repeats >= 1 times and matches "x*x": s[i - 1] == x && f[i - 1][j] * '.' matches any single character */ 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][n]; } };