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) {
for (int j = 1; j <= m; ++j) { // f[0][0]已经初始化过了,其余的f[i][0]一定不匹配(i>0),因此j从1开始
if (j + 1 <= m && p[j + 1] == '*') continue; // 例如a*要放在一起看
if (i && p[j] != '*') { // s[1]和p[1]存储的才是第一个有效字符
f[i][j] = f[i - 1][j - 1] && (s[i] == p[j] || p[j] == '.');
} else if (p[j] == '*') { // p[j]前面一定有有效字符,这是题目保证的
f[i][j] = f[i][j - 2] || (i && f[i - 1][j] && (s[i] == p[j - 1] || p[j - 1] == '.'));
}
}
}
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++)
for (int j = 1; j <= m; j++) {
if (j + 1 <= m && cp[j + 1] == '*') continue;
if (i != 0 && cp[j] != '*') {
f[i][j] = f[i - 1][j - 1] && (cs[i] == cp[j] || cp[j] == '.');
} else if (cp[j] == '*') {
f[i][j] = f[i][j - 2] || (i != 0 && f[i - 1][j] && (cs[i] == cp[j - 1] || cp[j - 1] == '.'));
}
}
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):
for j in range(1, m + 1):
if j + 1 <= m and p[j + 1] == '*':
continue
if i != 0 and p[j] != '*':
f[i][j] = f[i - 1][j - 1] and (s[i] == p[j] or p[j] == '.')
elif p[j] == '*':
f[i][j] = f[i][j - 2] or (i > 0 and f[i - 1][j] and (s[i] == p[j - 1] or p[j - 1] == '.'))
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:$O(n \times m)$。