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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 class Solution { 2 public: 3 bool isMatch(string s, string p) { 4 int m = s.size(), n = p.size(); 5 6 vector<vector<bool>> dp (m+1, vector<bool> (n+1, false)); 7 8 dp[0][0] = true; 9 10 for (int i = 0; i < m; i++){ 11 dp[i+1][0] = false; 12 } 13 14 for (int j = 0; j < n; j++){ 15 dp[0][j+1] = j > 0 && p[j] == ‘*‘ && dp[0][j-1]; 16 } 17 18 for (int i = 0; i < m; i++){ 19 for (int j = 0; j < n; j++){ 20 if (p[j] != ‘*‘){ 21 dp[i+1][j+1] = dp[i][j] && (s[i] == p[j] || p[j] == ‘.‘); 22 }else{ 23 dp[i+1][j+1] = dp[i+1][j-1] || ((dp[i][j+1]) && (s[i] == p[j-1] || p[j-1] == ‘.‘)); 24 } 25 } 26 } 27 28 return dp[m][n]; 29 } 30 };
[LeetCode #10] Regular Expression Matching
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原文地址：http://www.cnblogs.com/amadis/p/5926445.html
