Description#
Given an input string (s
) and a pattern (p
), implement wildcard pattern matching with support for '?'
and '*'
where:
'?'
Matches any single character.'*'
Matches any sequence of characters (including the empty sequence).
The matching should cover the entire input string (not partial).
Example 1:
Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
Example 2:
Input: s = "aa", p = "*"
Output: true
Explanation: '*' matches any sequence.
Example 3:
Input: s = "cb", p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.
Constraints:
0 <= s.length, p.length <= 2000
s
contains only lowercase English letters.p
contains only lowercase English letters, '?'
or '*'
.
Solutions#
Solution 1: Dynamic Programming#
We define the state $dp[i][j]$ to represent whether the first $i$ characters of $s$ match the first $j$ characters of $p$.
The state transition equation is as follows:
$$
dp[i][j]=
\begin{cases}
dp[i-1][j-1] & \text{if } s[i-1]=p[j-1] \text{ or } p[j-1]=\text{?} \
dp[i-1][j-1] \lor dp[i-1][j] \lor dp[i][j-1] & \text{if } p[j-1]=\text{*} \
\text{false} & \text{otherwise}
\end{cases}
$$
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the lengths of $s$ and $p$ respectively.
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| class Solution:
def isMatch(self, s: str, p: str) -> bool:
m, n = len(s), len(p)
dp = [[False] * (n + 1) for _ in range(m + 1)]
dp[0][0] = True
for j in range(1, n + 1):
if p[j - 1] == '*':
dp[0][j] = dp[0][j - 1]
for i in range(1, m + 1):
for j in range(1, n + 1):
if s[i - 1] == p[j - 1] or p[j - 1] == '?':
dp[i][j] = dp[i - 1][j - 1]
elif p[j - 1] == '*':
dp[i][j] = dp[i - 1][j] or dp[i][j - 1]
return dp[m][n]
|
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| class Solution {
public boolean isMatch(String s, String p) {
int m = s.length(), n = p.length();
boolean[][] dp = new boolean[m + 1][n + 1];
dp[0][0] = true;
for (int j = 1; j <= n; ++j) {
if (p.charAt(j - 1) == '*') {
dp[0][j] = dp[0][j - 1];
}
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '?') {
dp[i][j] = dp[i - 1][j - 1];
} else if (p.charAt(j - 1) == '*') {
dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
}
}
}
return dp[m][n];
}
}
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| 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));
dp[0][0] = true;
for (int j = 1; j <= n; ++j) {
if (p[j - 1] == '*') {
dp[0][j] = dp[0][j - 1];
}
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (s[i - 1] == p[j - 1] || p[j - 1] == '?') {
dp[i][j] = dp[i - 1][j - 1];
} else if (p[j - 1] == '*') {
dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
}
}
}
return dp[m][n];
}
};
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| func isMatch(s string, p string) bool {
m, n := len(s), len(p)
dp := make([][]bool, m+1)
for i := range dp {
dp[i] = make([]bool, n+1)
}
dp[0][0] = true
for j := 1; j <= n; j++ {
if p[j-1] == '*' {
dp[0][j] = dp[0][j-1]
}
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
if s[i-1] == p[j-1] || p[j-1] == '?' {
dp[i][j] = dp[i-1][j-1]
} else if p[j-1] == '*' {
dp[i][j] = dp[i-1][j] || dp[i][j-1]
}
}
}
return dp[m][n]
}
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| using System.Linq;
public class Solution {
public bool IsMatch(string s, string p) {
if (p.Count(ch => ch != '*') > s.Length)
{
return false;
}
bool[,] f = new bool[s.Length + 1, p.Length + 1];
bool[] d = new bool[s.Length + 1]; // d[i] means f[0, j] || f[1, j] || ... || f[i, j]
for (var j = 0; j <= p.Length; ++j)
{
d[0] = j == 0 ? true : d[0] && p[j - 1] == '*';
for (var i = 0; i <= s.Length; ++i)
{
if (j == 0)
{
f[i, j] = i == 0;
continue;
}
if (p[j - 1] == '*')
{
if (i > 0)
{
d[i] = f[i, j - 1] || d[i - 1];
}
f[i, j] = d[i];
}
else if (p[j - 1] == '?')
{
f[i, j] = i > 0 && f[i - 1, j - 1];
}
else
{
f[i, j] = i > 0 && f[i - 1, j - 1] && s[i - 1] == p[j - 1];
}
}
}
return f[s.Length, p.Length];
}
}
|