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

第一遍：

 public class Solution {
     public boolean isMatch(String s, String p) {
         if(p.length() == 0) return s.length() == 0;
         if(s.length() == 0) return p.length() == 0;
         if(p.charAt(0) == '?' || p.charAt(0) == s.charAt(0)) return isMatch(s.substring(1), p.substring(1));
         else if(p.charAt(0) == '*'){
             for(int i = 0; i < s.length(); i ++){
                 if(isMatch(s.substring(i), p.substring(1))) return true;
             }
             return false;
         }
         else return false;
     }
 }

Time Limit Exceeded

"abbabbbaabaaabbbbbabbabbabbbabbaaabbbababbabaaabbab", "*aabb***aa**a******aa*"

网上做法：

贪心的策略，能匹配就一直往后遍历，匹配不上了就看看前面有没有'*'来救救场，再从'*'后面接着试。

 public class Solution {
     public boolean isMatch(String s, String p) {
         int i = 0;
         int j = 0;
         int star = -1;
         int mark = -1;
         while (i < s.length()){
             if (j < p.length() && (p.charAt(j) == '?' || p.charAt(j) == s.charAt(i))) {
                 ++i;
                 ++j;
             } else if (j < p.length() && p.charAt(j) == '*') {
                 star = j++;
                 mark = i;
             } else if (star != -1) {
                 j = star + 1;
                 i = ++mark;
             } else {
                 return false;
             }
         }
         while (j < p.length() && p.charAt(j) == '*') {// i == s.length()
             ++j;
         }
         return j == p.length();
     }
 }

DP 解法： 但是会memory  limit exceeded：

 public class Solution {
     public boolean isMatch(String s, String p) {
         if(p == null || p.length() == 0) return s == null || s.length() == 0;
         if(s == null || s.length() == 0){
             return p == null || p.length() == 0 || (p.charAt(0) == '*' && isMatch(s, p.substring(1)));
         }
         int plen = p.length();
         int slen = s.length();
         boolean[][] dp = new boolean[plen][slen];
         if(p.charAt(0) == s.charAt(0) || p.charAt(0) == '?' || p.charAt(0) == '*') dp[0][0] = true;
         for(int i = 1; i < plen; i ++){
             if(p.charAt(i) == '*') dp[i][0] = dp[i - 1][0];
             else break;
         }
         for(int j = 1; j < slen; j ++){
             if(p.charAt(0) == '*') dp[0][j] = dp[0][j - 1];
         }
         for(int i = 1; i < plen; i ++){
             for(int j = 1; j < slen; j ++){
                 if(p.charAt(i) == '?' || p.charAt(i) == s.charAt(j)) dp[i][j] = dp[i - 1][j - 1];
                 else if(p.charAt(i) == '*'){
                     dp[i][j] = dp[i - 1][j] || dp[i - 1][j - 1] || dp[i][j - 1];
                 }else{
                     dp[i][j] = false;
                 }
             }
         }
         return dp[plen - 1][slen - 1];
     }
 }