Wildcard Matching

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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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "*") → trueisMatch("aa", "a*") → trueisMatch("ab", "?*") → trueisMatch("aab", "c*a*b") → false

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

思路1：二维动归（提交竟然空间超出错误）。 dp[i][j]代表p[0...i]是否匹配 s[0...j]。

如果p[i]！='*'，t[i][j] == true 当 t[i-1][j-1]==true &&（p[i]==s[j]||p[i]='.'）
如果p[i]=='*'，t[i][j]== true 当其中一个m使得 t[i-1][m]==true，where 0<=m<j.

思路2：二维状态压缩为一维，原理是一样的。

publicclassSolution {
 
   // MLE, the states can be compressed.
   publicbooleanisMatchOld(String s, String p) {
       if(s == null)
           returnp == null;
       if(p == null)
           returns == null;
 
       intm = p.length();
       intn = s.length();
 
       boolean[][] dp = newboolean[m + 1][n + 1];
       dp[0][0] = true;
       for(inti = 1; i <= m; i++) {
           if(p.charAt(i - 1) == '*') {
                intj = n + 1;
                for(intk = 0; k <= n; k++) {
                    if(dp[i - 1][k])
                        j = k;
                }
                for(; j <= n; j++)
                    dp[i][j] = true;
 
           }else{
                for(intj = 1; j <= n; j++) {
                    if(dp[i - 1][j - 1]&& (p.charAt(i - 1) == s.charAt(j - 1) || p.charAt(i - 1) == '?'))
                        dp[i][j] = true;
                }
           }
           dp[i][0] = dp[i - 1][0] && (p.charAt(i - 1) == '*');
       }
 
       returndp[m][n];
    }
 
   publicbooleanisMatch(String s, String p) {
       if(s == null)
           returnp == null;
       if(p == null)
           returns == null;
 
       intm = p.length();
       intn = s.length();
 
       intcount = 0;
       for(inti = 0; i < m; i++) {
           if(p.charAt(i) != '*')
                count++;
       }
       if(count > n)
           returnfalse;
 
       boolean[] dp = newboolean[n + 1];
       dp[0] = true;
       for(inti = 1; i <= m; i++) {
           if(p.charAt(i - 1) == '*') {
                intj = n+1;
                for(intk = 0; k <= n; k++) {
                    if(dp[k]){
                        j = k;
                        break;
                    }
                }
                for(; j <= n; j++)
                    dp[j] = true;
 
           }else{
                for(intj = n; j >= 1; j--) {
                    dp[j]= (dp[j - 1]&& (p.charAt(i - 1) == s.charAt(j - 1) || p.charAt(i - 1) == '?'));
                }
           }
           dp[0] = dp[0] && (p.charAt(i - 1) == '*');
       }
 
       returndp[n];
    }
 
   publicstaticvoid main(String[] args) {
       System.out.println(newSolution().isMatch("ab","*a"));
    }
}

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