[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") → 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：，，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';}};

努力爱一个人。付出，不一定会有收获；