[leetcode]44. Wildcard Matching

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本文探讨了两种解决正则表达式匹配问题的方法：动态规划和回溯法。通过实例详细解析了每种方法的实现过程，特别是动态规划中如何处理'*'和'?'等特殊字符，以及回溯法中如何利用全局变量进行状态回溯。

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这题的dp和第10题的solution类似。
第10题的dp
http://www.cnblogs.com/grandyang/p/4461713.html

Solution 1:dp

这里dp的涵义和第10题一样

class Solution {
   public boolean isMatch(String s, String p) {
    	if(s==null || p==null) return false;
    	int s_len = s.length();
    	int p_len = p.length();
    	
    	boolean[][] d = new boolean[s_len+1][p_len+1];
    	d[0][0] = true;
    	
    	//初始化，s为空的时候，出现*的情况，*前必须也true，*这里才能true
    	for (int i = 1; i <= p_len; i++) {
			if ( p.charAt(i-1) == '*') d[0][i] = d[0][i-1];
		}
    	
    	for (int i = 1; i <= s_len; i++) {
			for (int j = 1; j <= p_len; j++) {
				if(p.charAt(j-1) == '?') d[i][j] = d[i-1][j-1];
				if(p.charAt(j-1) == s.charAt(i-1)) d[i][j] = d[i-1][j-1];
				if(p.charAt(j-1) == '*') d[i][j] = d[i][j-1] || d[i-1][j];
				// *为空     //*匹配了一个或多个
			}
		}
    	
     
       
    	return d[s_len][p_len];
    }
}

Solution 2:回溯法

个人觉得不大好理解，我也写不出来

在这里插入代码片

> s: abcded
p: a*df
as we can see first we meet " * ", p = 1, star = 1, and match = 1, s = 1;
then p= 2 and he will be waiting there......and when s = 3 we have a match, great ! Then p = 3, s = 4, then they do not match ,so sad!
so p will go back to "p = 2"(he will be waiting for a while ) and as we have "match = 3" and match++, so now match = 4, does not match , then match = 5, also s = 5, match again!!
so in my opinion , the global variable "match " is to store the last index in "s" when we meet star in "p", because as you can see from my example even though we can match later , but that does not mean the current substring is the right answer , "match" give us a chance to move on and visit more characters in s ....hope that helps!

boolean comparison(String str, String pattern) {
        int s = 0, p = 0, match = 0, starIdx = -1;            
        while (s < str.length()){
            // 如果匹配的话，advancing both pointers
            if (p < pattern.length()  && (pattern.charAt(p) == '?' || str.charAt(s) == pattern.charAt(p))){
                s++;
                p++;
            }
            // staridx记最近出现的*的位子，p是*后的元素，p在这里等着
            else if (p < pattern.length() && pattern.charAt(p) == '*'){
                starIdx = p;
                match = s;
                p++;
            }
           // 如果不匹配，在p回到最近的一次*的后面一个，s和match前进
            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();
}