Description
Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:
'.' Matches any single character.'*' Matches zero or more of the preceding element.
Return a boolean indicating whether the matching covers 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 = "a*"
Output: true
Explanation: '*' means zero or more of the preceding 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 (.)".
Constraints:
1 <= s.length <= 201 <= p.length <= 20s contains only lowercase English letters.p contains only lowercase English letters, '.', and '*'.- It is guaranteed for each appearance of the character
'*', there will be a previous valid character to match.
Solutions
Solution 1: Memoization Search
We design a function dfs(i, j), which indicates whether the i-th character of s matches the j-th character of p. The answer is dfs(0, 0).
The calculation process of the function dfs(i, j) is as follows:
- If j has reached the end of p, then if i has also reached the end of s, the match is successful, otherwise, the match fails.
- If the next character of j is
'*', we can choose to match 0 s[i] characters, which is dfs(i, j + 2). If i \lt m and s[i] matches p[j], we can choose to match 1 s[i] character, which is dfs(i + 1, j).
- If the next character of j is not
'*', then if i \lt m and s[i] matches p[j], it is dfs(i + 1, j + 1). Otherwise, the match fails.
During the process, we can use memoization search to avoid repeated calculations.
The time complexity is O(m × n), and the space complexity is O(m × n). Here, m and n are the lengths of s and p respectively.
PythonJavaC++GoRustJavaScriptC#CPHP
class Solution:
def isMatch(self, s: str, p: str) -> bool:
@cache
def dfs(i, j):
if j >= n:
return i == m
if j + 1 < n and p[j + 1] == '*':
return dfs(i, j + 2) or (
i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j)
)
return i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j + 1)
m, n = len(s), len(p)
return dfs(0, 0)(code-box)
class Solution {
private Boolean[][] f;
private String s;
private String p;
private int m;
private int n;
public boolean isMatch(String s, String p) {
m = s.length();
n = p.length();
f = new Boolean[m + 1][n + 1];
this.s = s;
this.p = p;
return dfs(0, 0);
}
private boolean dfs(int i, int j) {
if (j >= n) {
return i == m;
}
if (f[i][j] != null) {
return f[i][j];
}
boolean res = false;
if (j + 1 < n && p.charAt(j + 1) == '*') {
res = dfs(i, j + 2)
|| (i < m && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '.') && dfs(i + 1, j));
} else {
res = i < m && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '.') && dfs(i + 1, j + 1);
}
return f[i][j] = res;
}
}(code-box)
class Solution {
public:
bool isMatch(string s, string p) {
int m = s.size(), n = p.size();
int f[m + 1][n + 1];
memset(f, 0, sizeof f);
function<bool(int, int)> dfs = [&](int i, int j) -> bool {
if (j >= n) {
return i == m;
}
if (f[i][j]) {
return f[i][j] == 1;
}
int res = -1;
if (j + 1 < n && p[j + 1] == '*') {
if (dfs(i, j + 2) or (i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j))) {
res = 1;
}
} else if (i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j + 1)) {
res = 1;
}
f[i][j] = res;
return res == 1;
};
return dfs(0, 0);
}
};(code-box)
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 j >= n {
return i == m
}
if f[i][j] != 0 {
return f[i][j] == 1
}
res := -1
if j+1 < n && p[j+1] == '*' {
if dfs(i, j+2) || (i < m && (s[i] == p[j] || p[j] == '.') && dfs(i+1, j)) {
res = 1
}
} else if i < m && (s[i] == p[j] || p[j] == '.') && dfs(i+1, j+1) {
res = 1
}
f[i][j] = res
return res == 1
}
return dfs(0, 0)
}(code-box)
impl Solution {
pub fn is_match(s: String, p: String) -> bool {
let (m, n) = (s.len(), p.len());
let mut f = vec![vec![0; n + 1]; m + 1];
fn dfs(
s: &Vec<char>,
p: &Vec<char>,
f: &mut Vec<Vec<i32>>,
i: usize,
j: usize,
m: usize,
n: usize,
) -> bool {
if j >= n {
return i == m;
}
if f[i][j] != 0 {
return f[i][j] == 1;
}
let mut res = -1;
if j + 1 < n && p[j + 1] == '*' {
if dfs(s, p, f, i, j + 2, m, n)
|| (i < m && (s[i] == p[j] || p[j] == '.') && dfs(s, p, f, i + 1, j, m, n))
{
res = 1;
}
} else if i < m && (s[i] == p[j] || p[j] == '.') && dfs(s, p, f, i + 1, j + 1, m, n) {
res = 1;
}
f[i][j] = res;
res == 1
}
dfs(
&s.chars().collect(),
&p.chars().collect(),
&mut f,
0,
0,
m,
n,
)
}
}(code-box)
/**
* @param {string} s
* @param {string} p
* @return {boolean}
*/
var isMatch = function (s, p) {
const m = s.length;
const n = p.length;
const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
const dfs = (i, j) => {
if (j >= n) {
return i === m;
}
if (f[i][j]) {
return f[i][j] === 1;
}
let res = -1;
if (j + 1 < n && p[j + 1] === '*') {
if (dfs(i, j + 2) || (i < m && (s[i] === p[j] || p[j] === '.') && dfs(i + 1, j))) {
res = 1;
}
} else if (i < m && (s[i] === p[j] || p[j] === '.') && dfs(i + 1, j + 1)) {
res = 1;
}
f[i][j] = res;
return res === 1;
};
return dfs(0, 0);
};(code-box)
public class Solution {
private string s;
private string p;
private int m;
private int n;
private int[,] f;
public bool IsMatch(string s, string p) {
m = s.Length;
n = p.Length;
f = new int[m + 1, n + 1];
this.s = s;
this.p = p;
return dfs(0, 0);
}
private bool dfs(int i, int j) {
if (j >= n) {
return i == m;
}
if (f[i, j] != 0) {
return f[i, j] == 1;
}
int res = -1;
if (j + 1 < n && p[j + 1] == '*') {
if (dfs(i, j + 2) || (i < m && (s[i] == p[j] || p[j] == '.') && dfs(i + 1, j))) {
res = 1;
}
} else if (i < m && (s[i] == p[j] || p[j] == '.') && dfs(i + 1, j + 1)) {
res = 1;
}
f[i, j] = res;
return res == 1;
}
}(code-box)
#define MAX_LEN 1000
char *ss, *pp;
int m, n;
int f[MAX_LEN + 1][MAX_LEN + 1];
bool dfs(int i, int j) {
if (j >= n) {
return i == m;
}
if (f[i][j] != 0) {
return f[i][j] == 1;
}
int res = -1;
if (j + 1 < n && pp[j + 1] == '*') {
if (dfs(i, j + 2) || (i < m && (ss[i] == pp[j] || pp[j] == '.') && dfs(i + 1, j))) {
res = 1;
}
} else if (i < m && (ss[i] == pp[j] || pp[j] == '.') && dfs(i + 1, j + 1)) {
res = 1;
}
f[i][j] = res;
return res == 1;
}
bool isMatch(char* s, char* p) {
ss = s;
pp = p;
m = strlen(s);
n = strlen(p);
memset(f, 0, sizeof(f));
return dfs(0, 0);
}(code-box)
class Solution {
/**
* @param String $s
* @param String $p
* @return Boolean
*/
function isMatch($s, $p) {
$m = strlen($s);
$n = strlen($p);
$f = array_fill(0, $m + 1, array_fill(0, $n + 1, 0));
$dfs = function ($i, $j) use (&$s, &$p, $m, $n, &$f, &$dfs) {
if ($j >= $n) {
return $i == $m;
}
if ($f[$i][$j] != 0) {
return $f[$i][$j] == 1;
}
$res = -1;
if ($j + 1 < $n && $p[$j + 1] == '*') {
if (
$dfs($i, $j + 2) ||
($i < $m && ($s[$i] == $p[$j] || $p[$j] == '.') && $dfs($i + 1, $j))
) {
$res = 1;
}
} elseif ($i < $m && ($s[$i] == $p[$j] || $p[$j] == '.') && $dfs($i + 1, $j + 1)) {
$res = 1;
}
$f[$i][$j] = $res;
return $res == 1;
};
return $dfs(0, 0);
}
}(code-box)
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. The answer is f[m][n]. Initialize f[0][0] = true, indicating that the empty string and the empty regular expression match.
Similar to Solution 1, we can discuss different cases.
- If p[j - 1] is
'*', we can choose to match 0 s[i - 1] characters, which is f[i][j] = f[i][j - 2]. If s[i - 1] matches p[j - 2], we can choose to match 1 s[i - 1] character, which is f[i][j] = f[i][j] \lor f[i - 1][j].
- If p[j - 1] is not
'*', then if s[i - 1] matches p[j - 1], it is f[i][j] = f[i - 1][j - 1]. Otherwise, the match fails.
The time complexity is O(m × n), and the space complexity is O(m × n). Here, m and n are the lengths of s and p respectively.
PythonJavaC++GoRustJavaScriptC#PHPC
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 i in range(m + 1):
for j in range(1, n + 1):
if p[j - 1] == "*":
f[i][j] = f[i][j - 2]
if i > 0 and (p[j - 2] == "." or s[i - 1] == p[j - 2]):
f[i][j] |= f[i - 1][j]
elif i > 0 and (p[j - 1] == "." or s[i - 1] == p[j - 1]):
f[i][j] = f[i - 1][j - 1]
return f[m][n](code-box)
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 i = 0; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (p.charAt(j - 1) == '*') {
f[i][j] = f[i][j - 2];
if (i > 0 && (p.charAt(j - 2) == '.' || p.charAt(j - 2) == s.charAt(i - 1))) {
f[i][j] |= f[i - 1][j];
}
} else if (i > 0
&& (p.charAt(j - 1) == '.' || p.charAt(j - 1) == s.charAt(i - 1))) {
f[i][j] = f[i - 1][j - 1];
}
}
}
return f[m][n];
}
}(code-box)
class Solution {
public:
bool isMatch(string s, string p) {
int m = s.size(), n = p.size();
bool f[m + 1][n + 1];
memset(f, false, sizeof f);
f[0][0] = true;
for (int i = 0; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (p[j - 1] == '*') {
f[i][j] = f[i][j - 2];
if (i && (p[j - 2] == '.' || p[j - 2] == s[i - 1])) {
f[i][j] |= f[i - 1][j];
}
} else if (i && (p[j - 1] == '.' || p[j - 1] == s[i - 1])) {
f[i][j] = f[i - 1][j - 1];
}
}
}
return f[m][n];
}
};(code-box)
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 i := 0; i <= m; i++ {
for j := 1; j <= n; j++ {
if p[j-1] == '*' {
f[i][j] = f[i][j-2]
if i > 0 && (p[j-2] == '.' || p[j-2] == s[i-1]) {
f[i][j] = f[i][j] || f[i-1][j]
}
} else if i > 0 && (p[j-1] == '.' || p[j-1] == s[i-1]) {
f[i][j] = f[i-1][j-1]
}
}
}
return f[m][n]
}(code-box)
impl Solution {
pub fn is_match(s: String, p: String) -> bool {
let m = s.len();
let n = p.len();
let mut f = vec![vec![false; n + 1]; m + 1];
f[0][0] = true;
let s: Vec<char> = s.chars().collect();
let p: Vec<char> = p.chars().collect();
for i in 0..=m {
for j in 1..=n {
if p[j - 1] == '*' {
f[i][j] = f[i][j - 2];
if i > 0 && (p[j - 2] == '.' || p[j - 2] == s[i - 1]) {
f[i][j] = f[i][j] || f[i - 1][j];
}
} else if i > 0 && (p[j - 1] == '.' || p[j - 1] == s[i - 1]) {
f[i][j] = f[i - 1][j - 1];
}
}
}
f[m][n]
}
}(code-box)
/**
* @param {string} s
* @param {string} p
* @return {boolean}
*/
var isMatch = function (s, p) {
const m = s.length;
const n = p.length;
const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(false));
f[0][0] = true;
for (let i = 0; i <= m; ++i) {
for (let j = 1; j <= n; ++j) {
if (p[j - 1] === '*') {
f[i][j] = f[i][j - 2];
if (i && (p[j - 2] === '.' || p[j - 2] === s[i - 1])) {
f[i][j] |= f[i - 1][j];
}
} else if (i && (p[j - 1] === '.' || p[j - 1] === s[i - 1])) {
f[i][j] = f[i - 1][j - 1];
}
}
}
return f[m][n];
};(code-box)
public class Solution {
public bool IsMatch(string s, string p) {
int m = s.Length, n = p.Length;
bool[,] f = new bool[m + 1, n + 1];
f[0, 0] = true;
for (int i = 0; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (p[j - 1] == '*') {
f[i, j] = f[i, j - 2];
if (i > 0 && (p[j - 2] == '.' || p[j - 2] == s[i - 1])) {
f[i, j] |= f[i - 1, j];
}
} else if (i > 0 && (p[j - 1] == '.' || p[j - 1] == s[i - 1])) {
f[i, j] = f[i - 1, j - 1];
}
}
}
return f[m, n];
}
}(code-box)
class Solution {
/**
* @param String $s
* @param String $p
* @return Boolean
*/
function isMatch($s, $p) {
$m = strlen($s);
$n = strlen($p);
$f = array_fill(0, $m + 1, array_fill(0, $n + 1, false));
$f[0][0] = true;
for ($i = 0; $i <= $m; $i++) {
for ($j = 1; $j <= $n; $j++) {
if ($p[$j - 1] == '*') {
$f[$i][$j] = $f[$i][$j - 2];
if ($i > 0 && ($p[$j - 2] == '.' || $p[$j - 2] == $s[$i - 1])) {
$f[$i][$j] = $f[$i][$j] || $f[$i - 1][$j];
}
} elseif ($i > 0 && ($p[$j - 1] == '.' || $p[$j - 1] == $s[$i - 1])) {
$f[$i][$j] = $f[$i - 1][$j - 1];
}
}
}
return $f[$m][$n];
}
}(code-box)
bool isMatch(char* s, char* p) {
int m = strlen(s), n = strlen(p);
bool f[m + 1][n + 1];
memset(f, 0, sizeof(f));
f[0][0] = true;
for (int i = 0; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (p[j - 1] == '*') {
f[i][j] = f[i][j - 2];
if (i > 0 && (p[j - 2] == '.' || p[j - 2] == s[i - 1])) {
f[i][j] = f[i][j] || f[i - 1][j];
}
} else if (i > 0 && (p[j - 1] == '.' || p[j - 1] == s[i - 1])) {
f[i][j] = f[i - 1][j - 1];
}
}
}
return f[m][n];
}(code-box)
