Given an input string (s) and a pattern (p), 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:
scould be empty and contains only lowercase lettersa-z.pcould be empty and contains only lowercase lettersa-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 preceding 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
题解：这个题如果用暴力的想法很难想清楚，因为对于.*的处理很难，但是如果我们理解.*可匹配也可不匹配这样一个性质就很容易联想递归或者DP的思路了：
这道题分的情况的要复杂一些，先给出递归的解法：
- 若p为空，且s也为空，返回true，反之返回false
- 若p的长度为1，且s长度也为1，且相同或是p为'.'则返回true，反之返回false
- 若p的第二个字符不为*，且此时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匹配
- 若p的第二个字符为*，s不为空且字符匹配，调用递归函数匹配s和去掉前两个字符的p，若匹配返回true，否则s去掉首字母
- 返回调用递归函数匹配s和去掉前两个字符的p的结果
由于我对递归结束的判断实在是太恶心了。。。。于是时间和空间都很慢，不过因为是最好理解的思路，所以还是把代码扔上来：
1 class Solution { 2 public: 3 int in(char ch){ 4 if(ch=='.') return 2; 5 else if(ch=='*') return 3; 6 else return 1; 7 } 8 bool isMatch(string s, string p) { 9 int lens = s.length();int lenp = p.length(); 10 if(lens==0 && lenp==0) { return 1;} 11 if(lens==1 && lenp==1 && s[0]==p[0]) {return 1;} 12 if(lens==0 && lenp==1) return 0; 13 if(lens!=0 && lenp==0) { return 0;} 14 if(lens==0){ 15 if((in(p[0])==2&&in(p[1])!=3)||(in(p[0])==2 &&in(p[1])!=3)) return 0; 16 if(in(p[1])==3) return isMatch(s,p.substr(2,lenp)); 17 } 18 if(in(p[0])==1&&in(p[1])!=3){ 19 if(lens==0||p[0]!=s[0]) { return 0;} 20 else return isMatch(s.substr(1,lens),p.substr(1,lenp)); 21 } 22 if(in(p[0])==2&&in(p[1])!=3){ 23 if(lens==0) { return 0;} 24 else return isMatch(s.substr(1,lens),p.substr(1,lenp)); 25 } 26 if(in(p[0])==1&&in(p[1])==3){ 27 if(p[0]!=s[0]) return isMatch(s,p.substr(2,lenp)); 28 else return max(isMatch(s.substr(1,lens),p),isMatch(s,p.substr(2,lenp))); 29 } 30 if(in(p[0])==2&&in(p[1])==3) 31 return max(isMatch(s.substr(1,lens),p),isMatch(s,p.substr(2,lenp))); 32 cout<<4<<endl; return 1; 33 } 34 };
后来在网上看到了大佬原来可以这么写
1 class Solution { 2 public: 3 bool isMatch(string s, string p) { 4 if (p.empty()) return s.empty(); 5 if (p.size() > 1 && p[1] == '*') { 6 return isMatch(s, p.substr(2)) || (!s.empty() && (s[0] == p[0] || p[0] == '.') && isMatch(s.substr(1), p)); 7 } else { 8 return !s.empty() && (s[0] == p[0] || p[0] == '.') && isMatch(s.substr(1), p.substr(1)); 9 } 10 } 11 };
我们也可以用DP来解，定义一个二维的DP数组，其中dp[i][j]表示s[0,i)和p[0,j)是否match，然后有下面三种情况(下面部分摘自：https://leetcode.com/problems/regular-expression-matching/discuss/5684/9-lines-16ms-c-dp-solutions-with-explanations)：
1. P[i][j] = P[i - 1][j - 1], if p[j - 1] != '*' && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
2. P[i][j] = P[i][j - 2], if p[j - 1] == '*' and the pattern repeats for 0 times;
3.
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.
1 class Solution { 2 public: 3 bool isMatch(string s, string p) { 4 int m = s.size(), n = p.size(); 5 vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false)); 6 dp[0][0] = true; 7 for (int i = 0; i <= m; ++i) { 8 for (int j = 1; j <= n; ++j) { 9 if (j > 1 && p[j - 1] == '*') { 10 dp[i][j] = dp[i][j - 2] || (i > 0 && (s[i - 1] == p[j - 2] || p[j - 2] == '.') && dp[i - 1][j]); 11 } else { 12 dp[i][j] = i > 0 && dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.'); 13 } 14 } 15 } 16 return dp[m][n]; 17 } 18 };