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
public class Solution { public boolean isMatch(String s, String p) { if (s == null || p == null) return false; boolean[][] dp = new boolean[s.length()+1][p.length()+1]; dp[0][0] = true; for (int i = 0; i < p.length(); i++) { if(p.charAt(i) == '*' && dp[0][i-1]) dp[0][i+1] = true; } for (int i = 0; i < s.length(); i++) { for (int j = 0; j < p.length(); j++) { if(p.charAt(j) == '.' || p.charAt(j) == s.charAt(i)) dp[i+1][j+1] = dp[i][j]; else if(p.charAt(j) == '*') { if(p.charAt(j-1) != s.charAt(i) && p.charAt(j-1) != '.') dp[i+1][j+1] = dp[i+1][j-1]; else dp[i+1][j+1] = (dp[i+1][j] || dp[i][j+1] || dp[i+1][j-1]); } } } return dp[s.length()][p.length()]; }}新闻热点
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