leetcode 44. 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", "c*a*b") → false

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又是一道正则匹配的问题，这次用的是非dp的思路，如果s和p的每一位是一一对应的（包括？）则去匹配下一位，如果p是*，那么就去匹配s和p+1，如果匹配失败则回退，s++。

public class Solution {
public boolean isMatch(String str, String pattern) {
int s = 0, p = 0, match = 0, starIdx = -1;
while (s < str.length()){
// advancing both pointers
if (p < pattern.length()  && (pattern.charAt(p) == '?' || str.charAt(s) == pattern.charAt(p))){
s++;
p++;
}
// * found, only advancing pattern pointer
else if (p < pattern.length() && pattern.charAt(p) == '*'){
starIdx = p;
match = s;
p++;
}
// last pattern pointer was *, advancing string pointer
else if (starIdx != -1){
p = starIdx + 1;
match++;
s = match;
}
//current pattern pointer is not star, last patter pointer was not *
//characters do not match
else return false;
}

//check for remaining characters in pattern
while (p < pattern.length() && pattern.charAt(p) == '*')
p++;

return p == pattern.length();
}
}

补充一种dp方法：

public class Solution {
public boolean isMatch(String s, String p) {
int m = s.length();
int n = p.length();
boolean[][] dp = new boolean[m + 1][n + 1];
dp[0][0] = true;
for (int j = 1; j <= n; j++)
dp[0][j] = dp[0][j - 1] && p.charAt(j - 1) == '*';
for(int i = 1; i <= m; i++) {
for(int j = 1; j <= n; j++) {
if(p.charAt(j - 1) == s.charAt(i - 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]
;
}
}