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:

• 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 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

 

思路一：递归思路，先匹配第一个字符。终止条件：p到达末尾而s没有达到，返回false。

dfs(i, j)表示从s的第i个字符和p的第j个字符匹配。

Top-down

 1 class Solution {
 2 public:
 3     bool isMatch(string s, string p) {
 4         if (p.empty())
 5             return s.empty();
 6         bool first_match = (!s.empty() && (s[0] == p[0] || p[0] == '.')); //判断s和p的第一个字符是否匹配
 7         if (p.length() >= 2 && p[1] == '*') { //如果p前两个字符是[a-z|.]*,则选择s的第一个字符是否与其匹配
 8             return isMatch(s, p.substr(2)) || (first_match && isMatch(s.substr(1), p)); //前面那种情况选择直接跳过
 9         } else {
10             return first_match && isMatch(s.substr(1), p.substr(1)); //p首字符不是[a-z|.]*这种情况
11         }
12     }
13 };

思路二：记忆化搜索

 1 class Solution {
 2 public:
 3     bool isMatch(string s, string p) {
 4         int lens = s.length(), lenp = p.length();
 5         vector<vector<int> > memo(lens + 1, vector<int>(lenp + 1, -1));
 6         return isMatch(s, p, 0, 0, memo);
 7     }
 8     bool isMatch(string s, string p, int i, int j, vector<vector<int> > &memo) {
 9         if (memo[i][j] != -1) {
10             return memo[i][j];
11         }
12         int ans;
13         if (j == p.length()) {
14             ans = (i == s.length());
15         } else {
16             bool first_match = (i < s.length() && (p[j] == '.' || s[i] == p[j]));
17             if (j + 1 < p.length() && p[j + 1] == '*') {
18                 ans = isMatch(s, p, i, j + 2, memo) || (first_match && isMatch(s, p, i + 1, j, memo));
19             } else {
20                 ans = first_match && isMatch(s, p, i + 1, j + 1, memo);
21             }
22         }
23         memo[i][j] = ans;
24         return ans;
25     }
26 };

思路三：动态规划

基于思路二，转化成迭代形式。dp[i][j]表示从下标i开始的s子串与从下标j开始的p子串是否匹配

 1 class Solution {
 2 public:
 3     bool isMatch(string s, string p) {
 4         int lens = s.length(), lenp = p.length();
 5         vector<vector<bool> > dp(lens + 1, vector<bool>(lenp + 1, false));
 6         dp[lens][lenp] = true;
 7         for (int i = lens; i >= 0; i--) {
 8             for (int j = lenp - 1; j >= 0; j--) {
 9                 bool first_match = (i < lens) && (p[j] == '.' || (s[i] == p[j]));
10                 if (j + 1 < lenp && p[j + 1] == '*') {
11                     dp[i][j] = dp[i][j + 2] || (first_match && dp[i + 1][j]);
12                 } else {
13                     dp[i][j] = first_match && dp[i + 1][j + 1];
14                 }
15             } 
16         }
17         return dp[0][0];
18     }
19 };

还有另一种形式的dp，dp[i][j]表示长度为i的s和长度为j的p是否匹配。

 1 class Solution {
 2 public:
 3     bool isMatch(string s, string p) {
 4         int lens = s.length(), lenp = p.length();
 5         vector<vector<bool> > dp(lens + 1, vector<bool>(lenp + 1, false));
 6         dp[0][0] = true;
 7         for (int i = 0; i <= lens; i++) {
 8             for (int j = 1; j <= lenp; j++) {
 9                 if (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 && (s[i - 1] == p[j - 1] || p[j - 1] == '.') && dp[i - 1][j - 1];
13                 }
14             }
15         }
16         return dp[lens][lenp];
17     }
18 };