标签：bsp color while not with sequence function net put
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
递归完美解答，然而超时了
public class Solution { public boolean isMatch(String s, String p) { if (s.length()==0){ if ( p.length()>0&&p.charAt(0)==‘*‘){ return isMatch(s,p.substring(1)); }else { return p.length() == 0; } } if (p.length()==0){ return s.length()==0; } if (p.charAt(0)==‘?‘){ return isMatch(s.substring(1),p.substring(1)); }else if(p.charAt(0)==‘*‘) { return isMatch(s.substring(1),p)||isMatch(s,p.substring(1)); }else { if (p.charAt(0)==s.charAt(0)){ return isMatch(s.substring(1),p.substring(1)); }else { return false; } } } }
所以考虑用回溯。
此博客讲解的很详细，http://blog.csdn.net/u012848330/article/details/52596618
int bp=-1; int bs=-1; int i=0; int j=0; while (i<s.length()){ if (j<p.length()&&p.charAt(j)==‘*‘){ j++; bp=j; bs=i; } else if (j<p.length()&&(p.charAt(j)==‘?‘||p.charAt(j)==s.charAt(i))){ i++; j++; }else if(bp!=-1){ j=bp; i=++bs; } else { break; } } while (j<p.length()&&p.charAt(j)==‘*‘){ j++; } return i==s.length()&&j==p.length();
标签：bsp color while not with sequence function net put
原文地址：http://www.cnblogs.com/sure0328/p/7155160.html
