题目链接
tag:
- Hard；
- DP；
question：
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 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 = "cab"
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 = "misisp*."
Output: false
思路：
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:
- P[i][j] = P[i - 1][j - 1], if p[j - 1] != '*' && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
- P[i][j] = P[i][j - 2], if p[j - 1] == '*' and the pattern repeats for 0 times;
- 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.
C++ 解法：
class Solution {
public:
bool isMatch(string s, string p) {
int m = s.size(), n = p.size();
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 (j > 1 && 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];
}
};
Python 解法：
class Solution:
def isMatch(self, s: str, p: str) -> bool:
dp = []
defaultRow = [False] + [False for _ in p]
dp.append(list(defaultRow))
dp[0][0] = True
for x in range(len(p)):
if p[x] == '*':
dp[0][x+1] = dp[0][x-1]
for i in range(len(s)):
dp.append(list(defaultRow))
for j in range(len(p)):
if p[j] != '*':
if (s[i] == p[j] or p[j] == '.') and dp[i][j]:
dp[i+1][j+1] = True
elif p[j] == '.':
if dp[i][j]:
dp[i+1][j+1] = True
else: # *
#匹配0次
if dp[i+1][j-1]:
dp[i+1][j+1] = True
#匹配1次
if dp[i+1][j]:
dp[i+1][j+1] = True
#匹配多次
if (s[i] == p[j-1] or p[j-1] == '.') and dp[i][j+1]:
dp[i+1][j+1] = True
return dp[-1][-1]
