# LeetCode10. Regular Expression Matching（动态规划）

## 2019 年 09 月 24 日 • 阅读: 1833 • LeetCode • 阅读设置

### LeetCode10. Regular Expression Matching（动态规划）

• Hard
• Accepted：342,997
• Submissions：1,334,578

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

### 链接

https://leetcode.com/problems/regular-expression-matching/

### 题解

1. if s[i-1] == p[j-1] || p[j-1] == '.': dp[i][j] = dp[i-1][j-1]
2. if p[j-1] == '*':
if p[j - 2] != '*' && p[j-2] != s[i-1]: dp[i][j] = dp[i][j-2] // 0a
else: dp[i][j] = dp[i][j-2]   // 0a
dp[i][j] = dp[i-1][j-2] // 1a
dp[i][j] = dp[i-1][j]   // 1+a

### 代码

Time SubmittedStatusRuntimeMemoryLanguage
7 hours agoAccepted4 ms8.2 MBcpp
class Solution {
public:
bool isMatch(string s, string p) {
int n = s.length(), m = p.length();
bool dp[n+1][m+1];
memset(dp, 0, sizeof dp);
dp[0][0] = true;
for (int j = 2; j < m + 1; j += 2)
{
if (p[j - 1] == '*' && dp[0][j - 2])
dp[0][j] = true;
}
for (int i = 1; i < n + 1; i++)
{
for (int j = 1; j < m + 1; 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 (p[j - 2] != s[i - 1] && p[j - 2] != '.')
dp[i][j] = dp[i][j - 2];
else
dp[i][j] = (dp[i - 1][j] || dp[i - 1][j - 2] || dp[i][j - 2]);
}
}
}
return dp[n][m];
}
};

The end.
2019年9月24日 星期二