Wildcard Matching
Implement wildcard pattern matching with support for '?' and '*'.
'?' Matches any single character. '*' Matches any sequence of characters (including the empty sequence).
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", "*") → true
- isMatch("aa", "a*") → true
- isMatch("ab", "?*") → true
- isMatch("aab", "cab") → false
https://leetcode.com/problems/wildcard-matching/
Solution I: adopt from Regular Expression Matching, TLE
public class Solution { public boolean isMatch(String s, String p) { if (s.length() == 0 && p.length() == 0) { return true; } if (p.length() == 0) { return false; } if (p.length() == 1) { if (s.length() == 0 || (s.charAt(0) != p.charAt(0) && p.charAt(0) != '?')) { return false; } if (p.charAt(0) == '') { return true; } return isMatch(s.substring(1), p.substring(1)); } char c = p.charAt(1); if (c != '') { if (s.length() == 0 || (s.charAt(0) != p.charAt(0) && p.charAt(0) != '?')) { return false; } return isMatch(s.substring(1), p.substring(1)); } else { if (isMatch(s, p.substring(2))) { return true; } int i = 0; while (i < s.length()) { if (isMatch(s.substring(i+1), p.substring(2))) { return true; } i++; } return false; } } }
Solution II: DP (2D)
https://leetcode.com/discuss/66038/java-solution-o-n-2-dp-solution-with-some-explanations
public class Solution { public boolean isMatch(String s, String p) { if(p.length()==0) return s.length()==0; boolean[][] res = new boolean[p.length()+1][s.length()+1]; res[0][0] = true; for (int i = 1; i <= p.length(); i++) { boolean flag = false; for (int j = 0; j <= s.length(); j++) { // note that j starts from 0 char c = p.charAt(i-1); flag = flag || res[i-1][j]; if (c != '') { res[i][j] = j>0 && res[i-1][j-1] && (c == '?' || s.charAt(j-1) == c); } else { // For k>=0 and k<=j, if any dp[i-1][k] is true, // then '' will match the rest sequence in s after index k; res[i][j] = i==1 || flag; // note that if i == 1 && c == '*', then res[1][j] = true } } } return res[p.length()][s.length()]; } }
Solution III: DP(1D)
http://blog.csdn.net/linhuanmars/article/details/21198049
public class Solution { public boolean isMatch(String s, String p) { if(p.length()==0) return s.length()==0; boolean[] res = new boolean[s.length()+1]; res[0] = true; for(int j=0;j