Wildcard Matching

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题目描述

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

这个题和前面的Regular Expression Matching差不多，我用了之前的dfs算法，超时了：

public boolean isMatch(String s, String p) {if(p.length()==0){return s.length()==0;}if(p.charAt(0)=='*'&&s.length()>0){if(isMatch(s, p.substring(1))){return true;}else{return isMatch(s.substring(1), p);}}if(s.length()==0){int start=0;while(start<p.length()&&p.charAt(start)=='*'){start++;}return start==p.length();}if(p.charAt(0)=='?'||(s.length()>0&&p.charAt(0)==s.charAt(0))){return isMatch(s.substring(1), p.substring(1));}return false;}

所以应该想到用动态规划算法，这里的isMatchEmpty(String s)是用来检测s是否能匹配空串，即"****"这样的字符串。

用dp[i][j]来保存s.subString(0,i)，与p.subString(0,j)是否匹配。

然后讨论当p.charAt(j)==’*‘，p.charAt(j)==’?‘，p.charAt(j)==s.charAt(i)，p.charAt(j)！=s.charAt(i)的情况，注意一些小细节~

代码如下：

public class Solution {    public boolean isMatch(String s, String p) {if(p.length()==0){return s.length()==0;}if(s.length()==0){return isMatchEmpty(p);}boolean[][] dp=new boolean[s.length()][p.length()];for(int i=0;i<s.length();i++){for(int j=0;j<p.length();j++){if(p.charAt(j)=='*'){if(i>0){if(j>0&&dp[i][j-1]){//这个不能掉dp[i][j]=true;}else{dp[i][j]=dp[i-1][j];}}else{if(j==0){dp[i][j]=true;}else{dp[i][j]=dp[0][j-1];}}}else if(p.charAt(j)=='?'||p.charAt(j)==s.charAt(i)){if(i>0&&j>0){dp[i][j]=dp[i-1][j-1];}else if(i==0&&j==0){dp[i][j]=true;}else if(i>0&&j==0){dp[i][j]=false;}else{dp[i][j]=isMatchEmpty(p.substring(0,j));}}else{dp[i][j]=false;}}}return dp[s.length()-1][p.length()-1];    }public boolean isMatchEmpty(String s){int i=0;while(i<s.length()&&s.charAt(i)=='*'){i++;}return i==s.length();}}

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