LeetCode – Regular Expression Matching (Java)
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") return false
isMatch("aa","aa") return true
isMatch("aaa","aa") return false
isMatch("aa", "a*") return true
isMatch("aa", ".*") return true
isMatch("ab", ".*") return true
isMatch("aab", "c*a*b") return true
1. Analysis
First of all, this is one of the most difficulty problems. It is hard to think through all different cases. The problem should be simplified to handle 2 basic cases:
- the second char of pattern is "*"
- the second char of pattern is not "*"
For the 1st case, if the first char of pattern is not ".", the first char of pattern and string should be the same. Then continue to match the remaining part.
For the 2nd case, if the first char of pattern is "." or first char of pattern == the first i char of string, continue to match the remaining part.
2. Java Solution 1 (Short)
The following Java solution is accepted.
public class Solution { public boolean isMatch(String s, String p) { if(p.length() == 0) return s.length() == 0; //p's length 1 is special case if(p.length() == 1 || p.charAt(1) != '*'){ if(s.length() < 1 || (p.charAt(0) != '.' && s.charAt(0) != p.charAt(0))) return false; return isMatch(s.substring(1), p.substring(1)); }else{ int len = s.length(); int i = -1; while(i<len && (i < 0 || p.charAt(0) == '.' || p.charAt(0) == s.charAt(i))){ if(isMatch(s.substring(i+1), p.substring(2))) return true; i++; } return false; } } } |
3. Java Solution 2 (More Readable)
public boolean isMatch(String s, String p) { // base case if (p.length() == 0) { return s.length() == 0; } // special case if (p.length() == 1) { // if the length of s is 0, return false if (s.length() < 1) { return false; } //if the first does not match, return false else if ((p.charAt(0) != s.charAt(0)) && (p.charAt(0) != '.')) { return false; } // otherwise, compare the rest of the string of s and p. else { return isMatch(s.substring(1), p.substring(1)); } } // case 1: when the second char of p is not '*' if (p.charAt(1) != '*') { if (s.length() < 1) { return false; } if ((p.charAt(0) != s.charAt(0)) && (p.charAt(0) != '.')) { return false; } else { return isMatch(s.substring(1), p.substring(1)); } } // case 2: when the second char of p is '*', complex case. else { //case 2.1: a char & '*' can stand for 0 element if (isMatch(s, p.substring(2))) { return true; } //case 2.2: a char & '*' can stand for 1 or more preceding element, //so try every sub string int i = 0; while (i<s.length() && (s.charAt(i)==p.charAt(0) || p.charAt(0)=='.')){ if (isMatch(s.substring(i + 1), p.substring(2))) { return true; } i++; } return false; } } |
Java Solution 3 - Using Index
We can also add a helper method to use the index, instead of passing sub-strings recursively.
public boolean isMatch(String s, String p) { return helper(s, p, 0, 0); } private boolean helper(String s, String p, int i, int j) { if (i >= s.length() && j >= p.length()) { return true; } if (j < p.length() - 1 && p.charAt(j + 1) == '*') { if (helper(s, p, i, j + 2)) { return true; } if (p.charAt(j) == '.') { for (int k = i; k < s.length(); k++) { if (helper(s, p, k + 1, j + 2)) { return true; } } } else { for (int k = i; k < s.length(); k++) { if (s.charAt(k) == p.charAt(j)) { if (helper(s, p, k + 1, j + 2)) { return true; } } else { break; } } } } else if (i < s.length() && j < p.length() && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '.')) { return helper(s, p, i + 1, j + 1); } return false; } |
<pre><code> String foo = "bar"; </code></pre>
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