[LeetCode] Wildcard Matching
Wildcard Matching
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
解題思路:
這道題與http://blog.csdn.net/kangrydotnet/article/details/46624353類似,註意這裏的*號與Regular Expression Matching不同。能夠採用Regular Expression Matching開發架構,用遞歸的方法。
可是會出現超時問題:
class Solution { public: bool isMatch(string s, string p) { return matchHelper(s, p, 0, 0); } bool matchHelper(string& s, string& p, int i, int j){ if(p[j]=='\0'){ return s[i]=='\0'; } if(p[j]!='*'){ return ((s[i]==p[j] || p[j]=='?'&&s[i]!='\0') && matchHelper(s, p, i+1, j+1)); } //p[j]=='*' while(s[i]!='\0'){ if(matchHelper(s, p, i, j+1)) return true; i++; } return matchHelper(s, p, i, j+1); } };
class Solution { public: bool isMatch(string s, string p) { string newP = ""; for (int i = 0; i < p.length(); i++){ if (i>0 && p[i - 1] == p[i] && p[i]=='*'){ continue; } newP += p[i]; } return matchHelper(s, newP, 0, 0); } bool matchHelper(string& s, string& p, int i, int j){ if (p[j] == '\0'){ return s[i] == '\0'; } if (p[j] != '*'){ return ((s[i] == p[j] || p[j] == '?'&&s[i] != '\0') && matchHelper(s, p, i + 1, j + 1)); } //p[j]=='*' while (s[i] != '\0'){ if (matchHelper(s, p, i, j + 1)) return true; i++; } return matchHelper(s, p, i, j + 1); } };
"abbabaaabbabbaababbabbbbbabbbabbbabaaaaababababbbabababaabbababaabbbbbbaaaabababbbaabbbbaabbbbababababbaabbaababaabbbababababbbbaaabbbbbabaaaabbababbbbaababaabbababbbbbababbbabaaaaaaaabbbbbaabaaababaaaabb", "**aa*****ba*a*bb**aa*ab****a*aaaaaa***a*aaaa**bbabb*b*b**aaaaaaaaa*a********ba*bbb***a*ba*bb*bb**a*b*bb"
經分析。遞歸調用時,會反復計算。能夠用一個二維數組d[i][j]記錄s[0...i]與p[0...j]是否匹配。
class Solution { public: bool isMatch(string s, string p) { string newP = ""; for (int i = 0; i < p.length(); i++){ if (i>0 && p[i - 1] == p[i] && p[i] == '*'){ continue; } newP += p[i]; } int sLen = s.length(); int pLen = newP.length(); vector<vector<bool>> d(sLen + 1, vector<bool>(pLen + 1, true)); return matchHelper(s, newP, 0, 0, d); } bool matchHelper(string& s, string& p, int i, int j, vector<vector<bool>>& d){ if (p[j] == '\0'){ return (d[i][j] = s[i] == '\0'); } if (p[j] != '*'){ return (d[i][j] = (s[i] == p[j] || p[j] == '?'&&s[i] != '\0') && d[i + 1][j + 1] && matchHelper(s, p, i + 1, j + 1, d)); } //p[j]=='*' while (s[i] != '\0'){ if ((d[i][j] = d[i][j + 1] && matchHelper(s, p, i, j + 1, d))) return true; i++; } return (d[i][j] = d[i][j + 1] && matchHelper(s, p, i, j + 1, d)); } };這裏運用短路原則。
速度倒是很快。可是會產生內存溢出錯誤。由於空間復雜度為O(m*n)。
終於辦法參考水中的魚:http://fisherlei.blogspot.com/2013/01/leetcode-wildcard-matching.html,詳細仍不是非常明確:
class Solution { public: bool isMatch(string s, string p) { bool star = false; int sStart = 0, pStart = 0; int str, ptr; for(str=sStart, ptr = pStart; s[str]!='\0'; str++, ptr++){ switch(p[ptr]){ case '?': break; case '*': star = true; sStart = str, pStart = ptr; while(p[pStart]=='*'){ pStart++; } if(p[pStart]=='\0'){ return true; } str = sStart - 1; ptr = pStart - 1; break; default: if(s[str]!=p[ptr]){ if(!star){ return false; } sStart++; str = sStart-1; ptr = pStart-1; } break; } } while(p[ptr]=='*') ptr++; return p[ptr]=='\0'; } };
[LeetCode] Wildcard Matching