[LeetCode123]Wildcard Matching
2014-07-01 06:54
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Implement wildcard pattern matching with support for
Analysis:
Greedy method
For each element in s
If *s==*p or *p == ? which means this is a match, then goes to next element s++ p++.
If p=='*', this is also a match, but one or many chars may be available, so let us save this *'s position and the matched s position.
If not match, then we check if there is a * previously showed up,
if there is no *, return false;
if there is an *, we set current p to the next element of *, and set current s to the next saved s position.
e.g.
abed
?b*d**
a=?, go on, b=b, go on,
e=*, save * position star=3, save s position ss = 3, p++
e!=d, check if there was a *, yes, ss++, s=ss; p=star+1
d=d, go on, meet the end.
check the rest element in p, if all are *, true, else false;
Note that in char array, the last is NOT NULL, to check the end, use "*p" or "*p=='\0'".
Java
c++
'?'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
Analysis:
Greedy method
For each element in s
If *s==*p or *p == ? which means this is a match, then goes to next element s++ p++.
If p=='*', this is also a match, but one or many chars may be available, so let us save this *'s position and the matched s position.
If not match, then we check if there is a * previously showed up,
if there is no *, return false;
if there is an *, we set current p to the next element of *, and set current s to the next saved s position.
e.g.
abed
?b*d**
a=?, go on, b=b, go on,
e=*, save * position star=3, save s position ss = 3, p++
e!=d, check if there was a *, yes, ss++, s=ss; p=star+1
d=d, go on, meet the end.
check the rest element in p, if all are *, true, else false;
Note that in char array, the last is NOT NULL, to check the end, use "*p" or "*p=='\0'".
Java
public boolean isMatch(String s, String p) { int slen = s.length(); int plen = p.length(); int i = 0, j=0; int star = -1; int sp = 0; while(i<slen){ while(j<plen && p.charAt(j)=='*'){ star = j++; sp = i; } if(j==plen || (s.charAt(i)!=p.charAt(j)&&p.charAt(j)!='?')){ if(star<0) return false; else { j = star+1; i = sp++; } }else { i++; j++; } } while(j<plen && p.charAt(j)=='*'){ j++; } return j == plen; }
c++
bool isMatch(const char *s, const char *p) { const char* star = NULL; const char* ss = s; while(*s){ if((*p=='?') || *p==*s){ s++; p++; continue; } if(*p=='*'){ star = p++; ss = s; continue; } if(star){ p = star+1;s=++ss; continue; } return false; } while(*p=='*'){p++;} return !*p; }
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