最近代碼寫的少了，而leetcode一直想做一個python，c/c++解題報告的專題，c/c++一直是我非常喜歡的，c語言編程練習的重要性體現在linux內核編程以及一些大公司算法上機的要求，python主要為了后序轉型數據分析和機器學習，所以今天來做一個難度為hard 的簡單正則表達式匹配。
做了很多leetcode題目，我們來總結一下套路： 首先一般是檢查輸入參數是否正確，然后是處理算法的特殊情況，之后就是實現邏輯，最后就是返回值。
當編程成為一種解決問題的習慣，我們就成為了一名純粹的程序員
（簡單正則表達式匹配）
題目描述
Implement regular expression matching with support for ‘.’ and ‘*’. ‘.’ Matches any single character. ‘*’ Matches zero or more of the preceding element. 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”, “a*”) → true isMatch(“aa”, “.*”) → true isMatch(“ab”, “.*”) → true isMatch(“aab”, “c*a*b”) → true
It might seem deceptively easy even you know the general idea, but programming it correctly with all the details require careful thought.
在程序設計中，規劃所有的細節問題都需要認真思考。
● It is a regular expression.Not a wild card.So the ” * ” does not mean any string.And the cab should be split like this “c * a * b” which means N “c”,N “a” and One “b”.
’ * ’ Matches zero or more of the preceding element, so ” c* ” could match nothing.
考慮情況：
s = “ac”, p = “ab*c”
After the first ‘a’, we see that there is no b’s to skip for “b*”. We match the last ‘c’ on both side and conclude that they both match.
It seems that being greedy is good. But how about this case?
s = “abbc”, p = “ab*bbc” When we see “b*” in p, we would have skip all b’s in s. They both should match, but we have no more b’s to match. Therefore, the greedy approach fails in the above case.
可能還有人說，如果碰到這種情況可以先看一下*后面的內容，但是碰見下面的情況就不好辦了。
s = “abcbcd”, p = “a.*c.*d” Here, “.*” in p means repeat ‘.’ 0 or more times. Since ‘.’ can match any character, it is not clear how many times ‘.’ should be repeated. Should the ‘c’ in p matches the first or second ‘c’ in s?
所以： Unfortunately, there is no way to tell without using some kind of exhaustive search（窮舉搜索）.
Hints:
A sample diagram of a deterministic finite state automata (DFA). DFAs are useful for doing lexical analysis and pattern matching. An example is UNIX’s grep tool. Please note that this post does not attempt to descibe a solution using DFA.
什么是DFA？
主要解決方案是回溯法，使用遞歸或者dp
We need some kind of backtracking mechanism （回溯法）such that when a matching fails, we return to the last successful matching state and attempt to match more characters in s with ‘*’. This approach leads naturally to recursion.
The recursion mainly breaks down elegantly to the following two cases: 主要考慮兩種遞歸情況 1. If the next character of p is NOT ‘*’, then it must match the current character of s. Continue pattern matching with the next character of both s and p. 2. If the next character of p is ‘*’, then we do a brute force exhaustive matching of 0, 1, or more repeats of current character of p… Until we could not match any more characters. You would need to consider the base case carefully too. That would be left as an exercise to the reader.
Below is the extremely concise code (Excluding comments and asserts, it’s about 10 lines of code).
1、考慮特殊情況即*s字符串或者*p字符串結束。 （1）s字符串結束，要求*p也結束或者間隔‘’ （例如p=”a*b*c……”），否則無法匹配 （2）*s字符串未結束，而*p字符串結束，則無法匹配 2、*s字符串與*p字符串均未結束 （1）(p+1)字符不為’‘，則只需比較s字符與*p字符，若相等則遞歸到(s+1)字符串與*(p+1)字符串的比較，否則無法匹配。 （2）(p+1)字符為’‘，則p字符可以匹配*s字符串中從0開始任意多（記為i）等于*p的字符，然后遞歸到(s+i+1)字符串與*(p+2)字符串的比較， 只要匹配一種情況就算完全匹配。
bool isMatch(const char *s,const char *p){ //判斷參數合法，以及程序正常結束 assert( s && p); if(*p == '/0') return *s == '/0'; //next char is not '*'; must match current character if(*(p+1) != '*') { assert(*p != '*');//考慮情況isMatch('aa','a*'); return ((*p == *s) ||(*p == '.' && *s != '/0')) && isMatch(s + 1, p + 1); } //next char is '*' 繼續遞歸匹配,不能寫成*(p+1) == '*' 考慮情況isMatch('ab','.*c') while((*p == *s)|| (*p == '.' && *s != '/0')) { if (isMatch(s, p+2)) return true; s++; } //匹配下一個模式 return isMatch(s,p+2);}此代碼運行時間：18ms 
Further Thoughts:
Some extra exercises to this problem: 1. If you think carefully, you can exploit some cases that the above code runs in exponential complexity. Could you think of some examples? How would you make the above code more efficient? 2. Try to implement partial matching instead of full matching. In addition, add ‘^’ and ‘$’ to the rule. ‘^’ matches the starting position within the string, while ‘$’ matches the ending position of the string. 3. Try to implement wildcard matching where ‘*’ means any sequence of zero or more characters. For the interested reader, real world regular expression matching (such as the grep tool) are usually implemented by applying formal language theory. To understand more about it, you may read this article. Rating: 4.8/5 (107 votes cast) Regular Expression Matching, 4.8 out of 5 based on 107 ratings
leetcode的 解題報告提醒我們說：
leetcode的解答報告中說的If you are stuck, recursion is your friend.
// 遞歸版，時間復雜度O(n)，空間復雜度O(1)
class Solution { public: bool isMatch(const char *s, const char *p) { if (*p == '/0') return *s == '/0'; // next char is not '*', then must match current character if (*(p + 1) != '*') { if (*p == *s || (*p == '.' && *s != '/0')) return isMatch(s + 1, p + 1); else return false; } else { // next char is '*' while (*p == *s || (*p == '.' && *s != '/0')) { if (isMatch(s, p + 2)) return true; s++; } return isMatch(s, p + 2); } }};My concise recursive and DP solutions with full explanation in C++ ● Please refer to my blog post if you have any comment. Wildcard matching problem can be solved similarly.
class Solution {public: bool isMatch(string s, string p) { if (p.empty()) return s.empty(); if ('*' == p[1]) // x* matches empty string or at least one character: x* -> xx* // *s is to ensure s is non-empty return (isMatch(s, p.substr(2)) || !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p)); else return !s.empty() && (s[0] == p[0] || '.' == p[0]) && isMatch(s.substr(1), p.substr(1)); }};class Solution {public: bool isMatch(string s, string p) { /** * f[i][j]: if s[0..i-1] matches p[0..j-1] * if p[j - 1] != '*' * f[i][j] = f[i - 1][j - 1] && s[i - 1] == p[j - 1] * if p[j - 1] == '*', denote p[j - 2] with x * f[i][j] is true iff any of the following is true * 1) "x*" repeats 0 time and matches empty: f[i][j - 2] * 2) "x*" repeats >= 1 times and matches "x*x": s[i - 1] == x && f[i - 1][j] * '.' matches any single character */ int m = s.size(), n = p.size(); vector<vector<bool>> f(m + 1, vector<bool>(n + 1, false)); f[0][0] = true; for (int i = 1; i <= m; i++) f[i][0] = false; // p[0.., j - 3, j - 2, j - 1] matches empty iff p[j - 1] is '*' and p[0..j - 3] matches empty for (int j = 1; j <= n; j++) f[0][j] = j > 1 && '*' == p[j - 1] && f[0][j - 2]; for (int i = 1; i <= m; i++) for (int j = 1; j <= n; j++) if (p[j - 1] != '*') f[i][j] = f[i - 1][j - 1] && (s[i - 1] == p[j - 1] || '.' == p[j - 1]); else // p[0] cannot be '*' so no need to check "j > 1" here f[i][j] = f[i][j - 2] || (s[i - 1] == p[j - 2] || '.' == p[j - 2]) && f[i - 1][j]; return f[m][n]; }};The shortest AC code.
1.’.’ is easy to handle. if p has a ‘.’, it can pass any single character in s except ‘/0’. 2.” is a totally different problem. if p has a ” character, it can pass any length of first-match characters in s including ‘/0’.
class Solution { public: bool matchFirst(const char *s, const char *p){ return (*p == *s || (*p == '.' && *s != '/0')); }bool isMatch(const char *s, const char *p) { if (*p == '/0') return *s == '/0'; //empty if (*(p + 1) != '*') {//without * if(!matchFirst(s,p)) return false; return isMatch(s + 1, p + 1); } else { //next: with a * if(isMatch(s, p + 2)) return true; //try the length of 0 while ( matchFirst(s,p) ) //try all possible lengths if (isMatch(++s, p + 2))return true; }}};a shorter one (14 lines of code) with neatly format:
class Solution {public: bool isMatch(const char *s, const char *p) { for( char c = *p; c != 0; ++s, c = *p ) { if( *(p+1) != '*' ) p++; else if( isMatch( s, p+2 ) ) return true; if( (*s==0) || ((c!='.') && (c!=*s)) ) return false; } return *s == 0; }};9-lines 16ms C++ DP Solutions with Explanations ●
This problem has a typical solution using Dynamic Programming. We define the state P[i][j] to be true if s[0..i) matches p[0..j) and false otherwise. Then the state equations are: a. P[i][j] = P[i - 1][j - 1], if p[j - 1] != ‘*’ && (s[i - 1] == p[j - 1] || p[j - 1] == ‘.’); b. P[i][j] = P[i][j - 2], if p[j - 1] == ‘*’ and the pattern repeats for 0 times; c. P[i][j] = P[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == ‘.’), if p[j - 1] == ‘*’ and the pattern repeats for at least 1 times. Putting these together, we will have the following code.
class Solution {public: bool isMatch(string s, string p) { int m = s.length(), n = p.length(); vector<vector<bool> > dp(m + 1, vector<bool> (n + 1, false)); dp[0][0] = true; for (int i = 0; i <= m; i++) for (int j = 1; j <= n; j++) if (p[j - 1] == '*') dp[i][j] = dp[i][j - 2] || (i > 0 && (s[i - 1] == p[j - 2] || p[j - 2] == '.') && dp[i - 1][j]); else dp[i][j] = i > 0 && dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.'); return dp[m][n]; }};2 years agoreply quote
使用re庫，叼炸天！
import reclass Solution: # @return a boolean def isMatch(self, s, p): return re.match('^' + p + '$', s) != None# debugs = Solution()print (s.isMatch("aaa", ".*")）Python DP solution in 36 ms
def isMatch(self, s, p): m = len(s) n = len(p) dp = [[True] + [False] * m] for i in xrange(n): dp.append([False]*(m+1)) for i in xrange(1, n + 1): x = p[i-1] if x == '*' and i > 1: dp[i][0] = dp[i-2][0] for j in xrange(1, m+1): if x == '*': dp[i][j] = dp[i-2][j] or dp[i-1][j] or (dp[i-1][j-1] and p[i-2] == s[j-1]) or (dp[i][j-1] and p[i-2]=='.') elif x == '.' or x == s[j-1]: dp[i][j] = dp[i-1][j-1] return dp[n][m]about a year agoreply quote class Solution(object): def isMatch(self, s, p, memo={("",""):True}): if not p and s: return False if not s and p: return set(p[1::2]) == {"*"} and not (len(p) % 2) if (s,p) in memo: return memo[s,p] char, exp, prev = s[-1], p[-1], 0 if len(p) < 2 else p[-2] memo[s,p] =/ (exp == '*' and ((prev in {char, '.'} and self.isMatch(s[:-1], p, memo)) or self.isMatch(s, p[:-2], memo)))/ or/ (exp in {char, '.'} and self.isMatch(s[:-1], p[:-1], memo)) return memo[s,p]# 445 / 445 test cases passed.# Status: Accepted# Runtime: 72 ms8ms backtracking solution C++
//regular expression matching//first solution: using recursive versionclass Solution {public: bool isMatch(string s, string p) { int m = s.length(), n = p.length(); return backtracking(s, m, p, n); } bool backtracking(string& s, int i, string& p, int j) { if (i == 0 && j == 0) return true; if (i != 0 && j == 0) return false; if (i == 0 && j != 0) { //in this case only p == "c*c*c*" this pattern can match null string if (p[j-1] == '*') { return backtracking(s, i, p, j-2); } return false; } //now both i and j are not null if (s[i-1] == p[j-1] || p[j-1] == '.') { return backtracking(s, i - 1, p, j - 1); } else if (p[j-1] == '*') { //two cases: determines on whether p[j-2] == s[i-1] //first p[j-2]* matches zero characters of p if (backtracking(s, i, p, j - 2)) return true; //second consider whether p[j-2] == s[i-1], if true, then s[i-1] is matched, move to backtracking(i - 1, j) if (p[j-2] == s[i-1] || p[j-2] == '.') { return backtracking(s, i - 1, p, j); } return false; } return false; }};c語言參考解決方案： 3ms C solution using O(mn) time and O(n) space
bool isMatch(char *s, char *p){ int i; int ls = strlen(s); int lp = strlen(p); bool* m = malloc((ls + 1) * sizeof(bool)); // init m[0] = true; for (i = 1; i <= ls; i++) { m[i] = false; } int ip; for (ip = 0; ip < lp; ip++) { if (ip + 1 < lp && p[ip + 1] == '*') { // do nothing } else if (p[ip] == '*') { char c = p[ip - 1]; for (i = 1; i <= ls; i++) { m[i] = m[i] || (m[i - 1] && (s[i - 1] == c || c == '.')); } } else { char c = p[ip]; for (i = ls; i > 0; i--) { m[i] = m[i - 1] && (s[i - 1] == c || c == '.'); } m[0] = false; } } bool ret = m[ls]; free(m); return ret;}簡短的代碼：
bool isMatch(char* s, char* p) { while (*s) { if (*p&&*(p+1)=='*') { if (!(*p==*s||*p=='.')) {p+=2;continue;} if (!isMatch(s,p+2)) {s++;continue;} else return true; } if (*p==*s||*p=='.') {s++;p++;continue;} return false; } while(*p&&*(p+1)=='*') p+=2; return !*p;}http://articles.leetcode.com/regular-expression-matching http://blog.csdn.net/gatieme/article/details/51049244 http://www.jianshu.com/p/85f3e5a9fcda
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