标签：
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
int helper(const string& s,const string& p,int i,int j){ if(j==p.size()&&i==s.size()) return 2; if(s.size()==i&&p[j]!=‘*‘) return 0; if(j==p.size()) return 1; if(p[j]==‘*‘){ if(j+1<p.size()&&p[j+1]==‘*‘) return helper(s,p,i,j+1); for(int k=0;k<=s.length()-i;k++){ int ret=helper(s,p,i+k,j+1); if(ret==0||ret==2) return ret; } } if(s[i]==p[j]||p[j]==‘?‘){ return helper(s,p,i+1,j+1); } return 1; } bool isMatch(string s, string p) { return helper(s,p,0,0)>1; }
class Solution { public: bool isMatch(string s, string p) { int lenS = s.size(), lenP = p.size(), star = lenP, markS = 0, i = 0, j = 0; while (i != lenS) { if (s[i] == p[j] || ‘?‘ == p[j]) { ++i; ++j; continue; } if (‘*‘ == p[j]) { markS = i; star = j++; continue; } if (star != lenP) { i = ++markS; j = star + 1; continue; } return false; } while (‘*‘ == p[j]) ++j; return j == lenP; } };
标签：
原文地址：http://www.cnblogs.com/wqkant/p/5386620.html
