Wildcard Matching

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") → 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

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();
	}
}