# Wildcard Matching

Implement wildcard pattern matching with support for '?' and '*'.

'?' Matches any single character. '*' Matches any sequence of characters (including the empty sequence).

The matching should cover the entire input string (not partial).

The function prototype should be: bool isMatch(const char s, const char p)

Some examples:

• isMatch("aa","a") → false
• isMatch("aa","aa") → true
• isMatch("aaa","aa") → false
• isMatch("aa", "*") → true
• isMatch("aa", "a*") → true
• isMatch("ab", "?*") → true
• isMatch("aab", "cab") → false

https://leetcode.com/problems/wildcard-matching/

Solution I: adopt from Regular Expression Matching, TLE

``````public class Solution {
public boolean isMatch(String s, String p) {
if (s.length() == 0 && p.length() == 0) {
return true;
}

if (p.length() == 0) {
return false;
}

if (p.length() == 1) {
if (s.length() == 0 || (s.charAt(0) != p.charAt(0) && p.charAt(0) != '?')) {
return false;
}
if (p.charAt(0) == '*') {
return true;
}
return isMatch(s.substring(1), p.substring(1));
}

char c = p.charAt(1);

if (c != '*') {
if (s.length() == 0 || (s.charAt(0) != p.charAt(0) && p.charAt(0) != '?')) {
return false;
}

return isMatch(s.substring(1), p.substring(1));
} else {
if (isMatch(s, p.substring(2))) {
return true;
}

int i = 0;

while (i < s.length()) {
if (isMatch(s.substring(i+1), p.substring(2))) {
return true;
}
i++;
}

return false;
}
}
}
``````

Solution II: DP (2D)

https://leetcode.com/discuss/66038/java-solution-o-n-2-dp-solution-with-some-explanations

``````public class Solution {
public boolean isMatch(String s, String p) {

if(p.length()==0)
return s.length()==0;

boolean[][] res = new boolean[p.length()+1][s.length()+1];

res[0][0] = true;

for (int i = 1; i <= p.length(); i++) {
boolean flag = false;
for (int j = 0; j <= s.length(); j++) { // note that j starts from 0
char c = p.charAt(i-1);
flag = flag || res[i-1][j];

if (c != '*') {
res[i][j] = j>0 && res[i-1][j-1] && (c == '?' || s.charAt(j-1) == c);
} else {
// For k>=0 and k<=j, if any dp[i-1][k] is true,
// then '*' will match the rest sequence in s after index k;
res[i][j] = i==1 || flag; // note that if i == 1 && c == '*', then res[1][j] = true
}
}
}

return res[p.length()][s.length()];

}
}
``````

Solution III: DP(1D)

http://blog.csdn.net/linhuanmars/article/details/21198049

``````public class Solution {
public boolean isMatch(String s, String p) {
if(p.length()==0)
return s.length()==0;
boolean[] res = new boolean[s.length()+1];
res[0] = true;
for(int j=0;j<p.length();j++)
{
if(p.charAt(j)!='*')
{
for(int i=s.length()-1;i>=0;i--)
{
res[i+1] = res[i]&&(p.charAt(j)=='?'||s.charAt(i)==p.charAt(j));
}
}
else
{
int i = 0;
while(i<=s.length() && !res[i])
i++;
for(;i<=s.length();i++)
{
res[i] = true;
}
}
res[0] = res[0]&&p.charAt(j)=='*';
}
return res[s.length()];
}
}
``````