[leetcode] 10. Regular Expression Matching

[leetcode] 10. Regular Expression Matching

Description

Given an input string (s) and a pattern §, 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).

Note:

s could be empty and contains only lowercase letters a-z. p could be empty and contains only lowercase letters a-z, and characters like . or *. Example 1:

Input:

s = "aa"p = "a"

Output:

false

Explanation:

"a" does not match the entire string "aa".

Example 2:

Input:

s = "aa"p = "a*"

Output:

true

Explanation:

'*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:

s = "ab"p = ".*"

Output:

true

Explanation:

".*" means "zero or more (*) of any character (.)".

Example 4:

Input:

s = "aab"p = "c*a*b"

Output:

true

Explanation:

c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:

s = "mississippi"p = "mis*is*p*."

Output:

false

分析

题目的意思是：

动态规划 dp[i][j]表示s[i-1]与p[j-1]是否匹配 如果 p[j-1] == s[i-1] || p[j-1] == ‘.’, 此时dp[i][j] = dp[i-1][j-1]; 如果 p[j-1] == ‘*’ 分两种情况: 1: 如果p[j-2] != s[i-1] && p[j-2] != ‘.’, 此时dp[i][j] = dp[i][j-2] //*前面字符匹配0次 2: 如果p[j-2] == s[i-1] || p[j-2] == ‘.’ 此时dp[i][j] = dp[i][j-2] // *前面字符匹配0次 或者 dp[i][j] = dp[i][j-1] // *前面字符匹配1次 或者 dp[i][j] = dp[i-1][j] // *前面字符匹配多次

这题用递归的话，大致思路如下：

若p为空，s也为空，则返回true，否则返回false；若p的长度为1，s的长度也为1，且相同或p为‘.’，则返回true，否则返回false；若p的第二个字符不为*，此时s为空，则返回false，否则判断首字符是否匹配，且从各自的第二个字符开始调用递归函数；若p的第二个字符为*，若s不空且字符匹配，调用递归函数s和去掉前两个字符的p，若匹配，则返回true，否则s去掉首字母；返回调用递归函数匹配s和去掉前两个字符p的结果。

代码一 递归版本

class Solution {public: bool isMatch(string s, string p) { if(p.empty()) return s.empty(); if(p.size()==1){ return s.size()==1&&(s[0]==p[0]||p[0]=='.'); } if(p[1]!='*'){ if(s.empty()) return false; return (s[0]==p[0]||p[0]=='.')&&isMatch(s.substr(1),p.substr(1)); } while(!s.empty()&&(s[0]==p[0]||p[0]=='.')){ if(isMatch(s,p.substr(2))) return true; s=s.substr(1); } return isMatch(s,p.substr(2)); }};

代码二 动态规划版本

class Solution {public: bool isMatch(const char *s, const char *p) { int m=strlen(s); int n=strlen(p); if(m==0&&n==0){ return true; } bool dp[m+1][n+1]; memset(dp,false,sizeof(dp)); dp[0][0]=true; for(int i=1;i<=n;i++){ if(p[i-1]=='*'){ dp[0][i]=dp[0][i-2]; } } for(int i=1;i<=m;i++){ for(int j=1;j<=n;j++){ if(s[i-1]==p[j-1]||p[j-1]=='.'){ dp[i][j]=dp[i-1][j-1]; }else if(p[j-1]=='*'){ if(j!=1&&p[j-2]!='.'&&s[i-1]!=p[j-2]){ dp[i][j]=dp[i][j-2]; }else{ dp[i][j]=dp[i][j-1]||dp[i-1][j]||dp[i][j-2]; } } } } return dp[m][n]; }};

参考文献

​​[编程题]regular-expression-matching​​​​[LeetCode] Regular Expression Matching 正则表达式匹配​​

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