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", "c*a*b") → false 解题思路一:参考之前写的http://www.cnblogs.com/tonyluis/p/4464059.html,可以很轻松的用递归写出代码,JAVA实现如下:
static public boolean isMatch(String s, String p) { if(s.length()==0){ for(int i=0;i<p.length();i++) if(p.charAt(i)!=‘*‘) return false; return true; } if (p.length() == 0) return s.length() == 0; else if (p.length() == 1) return p.charAt(0)==‘*‘||(s.length() == 1&& (p.charAt(0) == ‘?‘ || s.charAt(0) == p.charAt(0))); if(p.charAt(0)!=‘*‘){ if(p.charAt(0)!=s.charAt(0)&&p.charAt(0)!=‘?‘) return false; return isMatch(s.substring(1),p.substring(1)); } int index=0; while(index<p.length()){ if(p.charAt(index)==‘*‘) index++; else break; } if(index==p.length()) return true; p=p.substring(index); for(int i=0;i<s.length();i++){ if(isMatch(s.substring(i),p)) return true; } return false; }
结果Time Limit Exceeded!也就是说这种类似暴力枚举的算法肯定不能通过!
解题思路二:
用两个指针indexS和indexP遍历s和p,同时用两个指针starIndex和sPosition来记录*遍历过程中*最后一次出现的位置和对应的indexP的位置,一旦s和p不匹配,就回到starIndex的位置,同时sPosition+1,接着遍历,直到遍历完整个s为止,注意边界条件,JAVA实现如下:
static public boolean isMatch(String s, String p) { int indexS=0,indexP=0,starIndex=-2,sPosition=-2; while(indexS<s.length()){ if(starIndex==p.length()-1) return true; if(indexP>=p.length()){ if(starIndex<0) return false; indexP=starIndex+1; indexS=++sPosition; } if(s.charAt(indexS)==p.charAt(indexP)||p.charAt(indexP)==‘?‘){ indexS++; indexP++; } else if(p.charAt(indexP)==‘*‘){ starIndex=indexP++; sPosition=indexS; } else if(starIndex>=0){ indexP=starIndex+1; indexS=++sPosition; } else return false; } while(indexP<p.length()){ if(p.charAt(indexP)!=‘*‘) return false; indexP++; } return true; }
时间: 2024-11-05 14:57:25