标签：[] for tip course put style character string rac
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
Here are some conditions to figure out, then the logic can be very straightforward.
1, If p.charAt(j) == s.charAt(i) : dp[i][j] = dp[i-1][j-1];
2, If p.charAt(j) == ‘.‘ : dp[i][j] = dp[i-1][j-1];
3, If p.charAt(j) == ‘*‘:
here are two sub conditions:
1 if p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2] //in this case, a* only counts as empty
2 if p.charAt(j-1) == s.charAt(i) or p.charAt(j-1) == ‘.‘:
dp[i][j] = dp[i-1][j] //in this case, a* counts as multiple a (s.charAt(i) of course match with * since s.charAt(i)==p.charAt(j-1))
or dp[i][j] = dp[i][j-1] // in this case, a* counts as single a
or dp[i][j] = dp[i][j-2] // in this case, a* counts as empty
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) == ‘.‘) {
dp[i+1][j+1] = dp[i][j];
}
if (p.charAt(j) == s.charAt(i)) {
dp[i+1][j+1] = dp[i][j];
}
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()];
}
}
10 Regular Expression Matching
标签：[] for tip course put style character string rac
原文地址：http://www.cnblogs.com/apanda009/p/7357047.html
