44. Wildcard Matching
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 <= 2000scontains only lowercase English letters.pcontains only lowercase English letters,'?'or'*'.
Solutions
Solution 1: Memoization Search
We design a function $dfs(i, j)$, which represents whether the string $s$ starting from the $i$-th character matches the string $p$ starting from the $j$-th character. The answer is $dfs(0, 0)$.
The execution process of the function $dfs(i, j)$ is as follows:
If $i \geq \textit{len}(s)$, then $dfs(i, j)$ is true only when $j \geq \textit{len}(p)$ or $p[j] = '*'$ and $dfs(i, j + 1)$ is true.
If $j \geq \textit{len}(p)$, then $dfs(i, j)$ is false.
If $p[j] = '*'$, then $dfs(i, j)$ is true if and only if $dfs(i + 1, j)$ or $dfs(i + 1, j + 1)$ or $dfs(i, j + 1)$ is true.
Otherwise, $dfs(i, j)$ is true if and only if $p[j] = '?'$ or $s[i] = p[j]$ and $dfs(i + 1, j + 1)$ is true.
To avoid repeated calculations, we use the method of memoization search and store the result of $dfs(i, j)$ in a hash table.
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the lengths of the strings $s$ and $p$, respectively.
Python3
class Solution:
def isMatch(self, s: str, p: str) -> bool:
@cache
def dfs(i: int, j: int) -> bool:
if i >= len(s):
return j >= len(p) or (p[j] == "*" and dfs(i, j + 1))
if j >= len(p):
return False
if p[j] == "*":
return dfs(i + 1, j) or dfs(i + 1, j + 1) or dfs(i, j + 1)
return (p[j] == "?" or s[i] == p[j]) and dfs(i + 1, j + 1)
return dfs(0, 0)
Java
class Solution {
private Boolean[][] f;
private char[] s;
private char[] p;
private int m;
private int n;
public boolean isMatch(String s, String p) {
this.s = s.toCharArray();
this.p = p.toCharArray();
m = s.length();
n = p.length();
f = new Boolean[m][n];
return dfs(0, 0);
}
private boolean dfs(int i, int j) {
if (i >= m) {
return j >= n || (p[j] == '*' && dfs(i, j + 1));
}
if (j >= n) {
return false;
}
if (f[i][j] != null) {
return f[i][j];
}
if (p[j] == '*') {
f[i][j] = dfs(i + 1, j) || dfs(i + 1, j + 1) || dfs(i, j + 1);
} else {
f[i][j] = (p[j] == '?' || s[i] == p[j]) && dfs(i + 1, j + 1);
}
return f[i][j];
}
}
C++
class Solution {
public:
bool isMatch(string s, string p) {
int m = s.size(), n = p.size();
int f[m + 1][n + 1];
memset(f, -1, sizeof(f));
function<bool(int, int)> dfs = [&](int i, int j) {
if (i >= m) {
return j >= n || (p[j] == '*' && dfs(i, j + 1));
}
if (j >= n) {
return false;
}
if (f[i][j] != -1) {
return f[i][j] == 1;
}
if (p[j] == '*') {
f[i][j] = dfs(i + 1, j) || dfs(i, j + 1) ? 1 : 0;
} else {
f[i][j] = (p[j] == '?' || s[i] == p[j]) && dfs(i + 1, j + 1) ? 1 : 0;
}
return f[i][j] == 1;
};
return dfs(0, 0);
}
};
Go
func isMatch(s string, p string) bool {
m, n := len(s), len(p)
f := make([][]int, m+1)
for i := range f {
f[i] = make([]int, n+1)
}
var dfs func(i, j int) bool
dfs = func(i, j int) bool {
if i >= m {
return j >= n || p[j] == '*' && dfs(i, j+1)
}
if j >= n {
return false
}
if f[i][j] != 0 {
return f[i][j] == 1
}
f[i][j] = 2
ok := false
if p[j] == '*' {
ok = dfs(i+1, j) || dfs(i+1, j+1) || dfs(i, j+1)
} else {
ok = (p[j] == '?' || s[i] == p[j]) && dfs(i+1, j+1)
}
if ok {
f[i][j] = 1
}
return ok
}
return dfs(0, 0)
}
TypeScript
function isMatch(s: string, p: string): boolean {
const m = s.length;
const n = p.length;
const f: number[][] = Array.from({ length: m + 1 }, () =>
Array.from({ length: n + 1 }, () => -1),
);
const dfs = (i: number, j: number): boolean => {
if (i >= m) {
return j >= n || (p[j] === '*' && dfs(i, j + 1));
}
if (j >= n) {
return false;
}
if (f[i][j] !== -1) {
return f[i][j] === 1;
}
if (p[j] === '*') {
f[i][j] = dfs(i + 1, j) || dfs(i, j + 1) ? 1 : 0;
} else {
f[i][j] = (p[j] === '?' || s[i] === p[j]) && dfs(i + 1, j + 1) ? 1 : 0;
}
return f[i][j] === 1;
};
return dfs(0, 0);
}
C#
public class Solution {
private bool?[,] f;
private char[] s;
private char[] p;
private int m;
private int n;
public bool IsMatch(string s, string p) {
this.s = s.ToCharArray();
this.p = p.ToCharArray();
m = s.Length;
n = p.Length;
f = new bool?[m, n];
return Dfs(0, 0);
}
private bool Dfs(int i, int j) {
if (i >= m) {
return j >= n || (p[j] == '*' && Dfs(i, j + 1));
}
if (j >= n) {
return false;
}
if (f[i, j] != null) {
return f[i, j].Value;
}
if (p[j] == '*') {
f[i, j] = Dfs(i + 1, j) || Dfs(i + 1, j + 1) || Dfs(i, j + 1);
} else {
f[i, j] = (p[j] == '?' || s[i] == p[j]) && Dfs(i + 1, j + 1);
}
return f[i, j].Value;
}
}
Solution 2: Dynamic Programming
We can convert the memoization search in Solution 1 into dynamic programming.
Define $f[i][j]$ to represent whether the first $i$ characters of string $s$ match the first $j$ characters of string $p$. Initially, $f[0][0] = \textit{true}$, indicating that two empty strings are matching. For $j \in [1, n]$, if $p[j-1] = '*'$, then $f[0][j] = f[0][j-1]$.
Next, we consider the case of $i \in [1, m]$ and $j \in [1, n]$:
If $p[j-1] = '*'$, then $f[i][j] = f[i-1][j] \lor f[i][j-1] \lor f[i-1][j-1]$.
Otherwise, $f[i][j] = (p[j-1] = '?' \lor s[i-1] = p[j-1]) \land f[i-1][j-1]$.
The final answer is $f[m][n]$.
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the lengths of the strings $s$ and $p$, respectively.
Python3
class Solution:
def isMatch(self, s: str, p: str) -> bool:
m, n = len(s), len(p)
f = [[False] * (n + 1) for _ in range(m + 1)]
f[0][0] = True
for j in range(1, n + 1):
if p[j - 1] == "*":
f[0][j] = f[0][j - 1]
for i in range(1, m + 1):
for j in range(1, n + 1):
if p[j - 1] == "*":
f[i][j] = f[i - 1][j] or f[i][j - 1] or f[i - 1][j - 1]
else:
f[i][j] = f[i - 1][j - 1] and (
p[j - 1] == "?" or s[i - 1] == p[j - 1]
)
return f[m][n]
Java
class Solution {
public boolean isMatch(String s, String p) {
int m = s.length(), n = p.length();
boolean[][] f = new boolean[m + 1][n + 1];
f[0][0] = true;
for (int j = 1; j <= n; ++j) {
if (p.charAt(j - 1) == '*') {
f[0][j] = f[0][j - 1];
}
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (p.charAt(j - 1) == '*') {
f[i][j] = f[i - 1][j] || f[i][j - 1] || f[i - 1][j - 1];
} else {
f[i][j] = f[i - 1][j - 1]
&& (p.charAt(j - 1) == '?' || s.charAt(i - 1) == p.charAt(j - 1));
}
}
}
return f[m][n];
}
}
C++
class Solution {
public:
bool isMatch(string s, string p) {
int m = s.length(), n = p.length();
bool f[m + 1][n + 1];
memset(f, false, sizeof(f));
f[0][0] = true;
for (int j = 1; j <= n; ++j) {
if (p[j - 1] == '*') {
f[0][j] = f[0][j - 1];
}
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (p[j - 1] == '*') {
f[i][j] = f[i - 1][j] || f[i][j - 1] || f[i - 1][j - 1];
} else {
f[i][j] = f[i - 1][j - 1] && (p[j - 1] == '?' || s[i - 1] == p[j - 1]);
}
}
}
return f[m][n];
}
};
Go
func isMatch(s string, p string) bool {
m, n := len(s), len(p)
f := make([][]bool, m+1)
for i := range f {
f[i] = make([]bool, n+1)
}
f[0][0] = true
for j := 1; j <= n; j++ {
if p[j-1] == '*' {
f[0][j] = f[0][j-1]
}
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
if p[j-1] == '*' {
f[i][j] = f[i-1][j] || f[i][j-1] || f[i-1][j-1]
} else {
f[i][j] = f[i-1][j-1] && (p[j-1] == '?' || s[i-1] == p[j-1])
}
}
}
return f[m][n]
}
TypeScript
function isMatch(s: string, p: string): boolean {
const m: number = s.length;
const n: number = p.length;
const f: boolean[][] = Array.from({ length: m + 1 }, () =>
Array.from({ length: n + 1 }, () => false),
);
f[0][0] = true;
for (let j = 1; j <= n; ++j) {
if (p.charAt(j - 1) === '*') {
f[0][j] = f[0][j - 1];
}
}
for (let i = 1; i <= m; ++i) {
for (let j = 1; j <= n; ++j) {
if (p[j - 1] === '*') {
f[i][j] = f[i - 1][j] || f[i][j - 1] || f[i - 1][j - 1];
} else {
f[i][j] = f[i - 1][j - 1] && (p[j - 1] === '?' || s[i - 1] === p[j - 1]);
}
}
}
return f[m][n];
}
PHP
class Solution {
/**
* @param string $s
* @param string $p
* @return boolean
*/
function isMatch($s, $p) {
$lengthS = strlen($s);
$lengthP = strlen($p);
$dp = [];
for ($i = 0; $i <= $lengthS; $i++) {
$dp[$i] = array_fill(0, $lengthP + 1, false);
}
$dp[0][0] = true;
for ($i = 1; $i <= $lengthP; $i++) {
if ($p[$i - 1] == '*') {
$dp[0][$i] = $dp[0][$i - 1];
}
}
for ($i = 1; $i <= $lengthS; $i++) {
for ($j = 1; $j <= $lengthP; $j++) {
if ($p[$j - 1] == '?' || $s[$i - 1] == $p[$j - 1]) {
$dp[$i][$j] = $dp[$i - 1][$j - 1];
} elseif ($p[$j - 1] == '*') {
$dp[$i][$j] = $dp[$i][$j - 1] || $dp[$i - 1][$j];
}
}
}
return $dp[$lengthS][$lengthP];
}
}