[LeetCode] Wildcard Matching
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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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false
解题思路:
这道题与http://blog.csdn.net/kangrydotnet/article/details/46624353类似,注意这里的*号与Regular Expression Matching不同。可以采用Regular Expression Matching开发架构,用递归的方法。但是会出现超时问题:
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); }};当输入"aaabbbaabaaaaababaabaaabbabbbbbbbbaabababbabbbaaaaba", "a*******b"时,上述代码超时。原因是有连续的*。于是修正输入,将a******b改成a*b,如下:
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);}};当输入非常大的时候,仍然超时:
"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"
经分析,递归调用时,会重复计算。可以用一个二维数组d[i][j]记录s[0...i]与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));}};这里运用短路原则。速度倒是非常快,但是会产生内存溢出错误。因为空间复杂度为O(m*n)。
最终办法参考水中的鱼:http://fisherlei.blogspot.com/2013/01/leetcode-wildcard-matching.html,具体仍不是很明白:
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';}};
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