Leetcode: Regular Expression Matching 这道每次都要看答案 心塞 啊啊啊
2015-02-15 09:10
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Problem:
Implement regular expression matching with support for
Solution:
Code:
public class Solution {
public boolean isMatch(String s, String p) {
if (p.length() == 0) {
if (s.length() == 0) return true;
else return false;
}
if (p.length() == 1 || p.charAt(1) != '*') {
if (s.length() == 0 || (s.charAt(0) != p.charAt(0) && p.charAt(0) != '.')) {
return false;
} else {
return isMatch(s.substring(1), p.substring(1));
}
} else {
int i = -1;
while(i < s.length() && (i == -1 || ( (s.charAt(i) == p.charAt(0)) || p.charAt(0) == '.'))) {
i++;
if (isMatch(s.substring(i), p.substring(2))) return true;
}
}
return false;
}
}
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
Solution:
Code:
public class Solution {
public boolean isMatch(String s, String p) {
if (p.length() == 0) {
if (s.length() == 0) return true;
else return false;
}
if (p.length() == 1 || p.charAt(1) != '*') {
if (s.length() == 0 || (s.charAt(0) != p.charAt(0) && p.charAt(0) != '.')) {
return false;
} else {
return isMatch(s.substring(1), p.substring(1));
}
} else {
int i = -1;
while(i < s.length() && (i == -1 || ( (s.charAt(i) == p.charAt(0)) || p.charAt(0) == '.'))) {
i++;
if (isMatch(s.substring(i), p.substring(2))) return true;
}
}
return false;
}
}
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