[LintCode] Wildcard Matching

时间：2017-08-29 12:59:04 阅读：169 评论：0 收藏：0 [点我收藏+]

标签：express port bool mic class null ble and redundant

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).

Example

isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false

Solution 1. Recursion

 1 public class WildcardMatching {
 2     public boolean isMatchRecursion(String str, String pattern) {
 3         if(str == null || pattern == null) {
 4             return false;
 5         }
 6         String newPattern = pattern.replaceAll("(\\*){2,}", "*");       
 7         return matchHelper(str, str.length() - 1, newPattern, newPattern.length() - 1);
 8     }
 9     private boolean matchHelper(String str, int idx1, String pattern, int idx2) {
10         if((idx1 < 0 && idx2 < 0) || (idx1 < 0 && idx2 == 0 && pattern.charAt(idx2) == ‘*‘)) {
11             return true;
12         }
13         else if(idx1 < 0 && idx2 >= 0 || idx1 >= 0 && idx2 < 0) {
14             return false;
15         }
16         if(pattern.charAt(idx2) == ‘?‘ || str.charAt(idx1) == pattern.charAt(idx2)) {
17             return matchHelper(str, idx1 - 1, pattern, idx2 - 1);
18         }
19         else if(pattern.charAt(idx2) == ‘*‘) {
20             return matchHelper(str, idx1, pattern, idx2 - 1) || matchHelper(str, idx1 - 1, pattern, idx2);
21         }
22         return false;
23     }
24     private String removeRedundantStars(String s) {
25         char[] chars = s.toCharArray();
26         int j = chars.length - 1, i = chars.length - 1;
27         while(j >= 0) {
28             chars[i] = chars[j];
29             if(chars[j] == ‘*‘) {
30                 while(j >= 0 && chars[j] == ‘*‘){
31                     j--;
32                 }                
33             }
34             else {
35                 j--;
36             }
37             i--;
38         }
39         return String.copyValueOf(chars, i + 1, chars.length - i - 1);
40     }
41 }

Solution 2. Dynamic Programming

 1 public boolean isMatchDp(String str, String pattern) {
 2     if(str == null || pattern == null) {
 3         return false;
 4     }
 5     String newPattern = pattern.replaceAll("(\\*){2,}", "*");    
 6     boolean[][] T = new boolean[str.length() + 1][newPattern.length() + 1];
 7     T[0][0] = true;
 8     if(newPattern.length() > 0 && newPattern.charAt(0) == ‘*‘) {
 9         T[0][1] = true;
10     }
11     for(int i = 1; i <= str.length(); i++) {
12         for(int j = 1; j <= newPattern.length(); j++) {
13             if(str.charAt(i - 1) == newPattern.charAt(j - 1) || newPattern.charAt(j - 1) == ‘?‘) {
14                 T[i][j] = T[i - 1][j - 1];
15             }
16             else if(newPattern.charAt(j - 1) == ‘*‘) {
17                 T[i][j] = T[i][j - 1] || T[i - 1][j];
18             }
19             else {
20                 T[i][j] = false;
21             }
22         }
23     }
24     return T[str.length()][newPattern.length()];
25 }