分析
-
本题的考点:动态规划。
-
分析如下:
代码
- C++
class Solution {
public:
bool isMatch(string s, string p) {
int n = s.size(), m = p.size();
s = " " + s, p = ' ' + p;
vector<vector<bool>> f(n + 1, vector<bool>(m + 1));
f[0][0] = true;
for (int i = 0; i <= n; i++) // f[0][j]是有意义的,例如p="***",因此i从0开始
for (int j = 1; j <= m; j++) // f[i][0] (i>0)是没有意义的,因此j从1开始
if (p[j] != '*') {
f[i][j] = (s[i] == p[j] || p[j] == '?') && i && f[i - 1][j - 1];
} else {
f[i][j] = f[i][j - 1] || i && f[i - 1][j];
}
return f[n][m];
}
};
- Java
class Solution {
public boolean isMatch(String s, String p) {
int n = s.length(), m = p.length();
char[] cs = (" " + s).toCharArray(), cp = (' ' + p).toCharArray();
boolean[][] f = new boolean[n + 1][m + 1];
f[0][0] = true;
for (int i = 0; i <= n; i++) // f[0][j]是有意义的,例如p="***",因此i从0开始
for (int j = 1; j <= m; j++) // f[i][0] (i>0)是没有意义的,因此j从1开始
if (cp[j] != '*') {
f[i][j] = (cs[i] == cp[j] || cp[j] == '?') && i > 0 && f[i - 1][j - 1];
} else {
f[i][j] = f[i][j - 1] || i > 0 && f[i - 1][j];
}
return f[n][m];
}
}
- Python
class Solution:
def isMatch(self, s: str, p: str) -> bool:
n = len(s); m = len(p)
s = " " + s; p = ' ' + p
f = [[False for _ in range(m + 1)] for _ in range(n + 1)]
f[0][0] = True
for i in range(n + 1): # f[0][j]是有意义的,例如p="***",因此i从0开始
for j in range(1, m + 1): # f[i][0] (i>0)是没有意义的,因此j从1开始
if p[j] != '*':
f[i][j] = (s[i] == p[j] or p[j] == '?') and i > 0 and f[i - 1][j - 1]
else:
f[i][j] = f[i][j - 1] or (i > 0 and f[i - 1][j])
return f[n][m]
时空复杂度分析
-
时间复杂度:$O(n \times m)$,
n、m
分别为s、p
的长度。 -
空间复杂度:$O(n \times m)$。