LeetCode 044.Wildcard Matching

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本文深入探讨了通配符匹配算法，特别是针对'?'和'*'的动态规划解决方案。通过多个实例，详细解释了如何使用List[i][j]表示字符串s与p的匹配状态，以及当遇到'*'时的两种不同处理思路。最终，提供了简洁高效的Java代码实现。

Given an input string (s) and a pattern (p), 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).

Note:

s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like ? or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "*"
Output: true
Explanation: '*' matches any sequence.

Example 3:

Input:
s = "cb"
p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.

Example 4:

Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".

Example 5:

Input:
s = "acdcb"
p = "a*c?b"
Output: false

/*********************************************************/

又是一道我用naive方法做了很久，搞了半天，最后超时限。回过头来用动态规划做，一下就做完了的题目……

- 这道题动态规划的重点在于：用List[i][j]表示字符串s从0到i与字符串p从0到j进行匹配时的结果。当字符串p中j的位置是正确的字母或是？时，都非常简单，直接赋值List[i-1][j-1]即可。但如果p中j位置上是*，那么就变得复杂了。一种思路认为，当j位置上为*时

List[i][j] = List[i][j-1] || List[0][j-1] || List[1][j-1] || ........ || List[i-1][j-1]

因为当j位是*时，它可能不匹配，也可能匹配一个字符，也可能匹配n(0<=n<=i)个字符。

另一种思路则更简单——当j位置上为*时

List[i][j] = List[i][j-1] || List[i-1][j]

因为当*只要匹配一个以上的字符时，List[i][j] 的结果与 List[i-1][j] 一定是一样的。

有了思路代码就非常简单了，放上我自己的代码，语言是java

class Solution {
    public boolean isMatch(String s, String p) {
		int ls = s.length();
		int lp = p.length();
		boolean List[][] = new boolean[ls+1][lp+1];
		for(int k = 1;k<=ls;k++)
			List[k][0] = false;
		List[0][0] = true;
        for(int k = 0;k<lp && p.charAt(k) == '*' ;k++)
        {
            List[0][k+1] = true;
        }
		for(int j = 0; j<lp;j++)
		{
			for(int i = 0;i<ls;i++)
			{
				if(p.charAt(j)>='a' && p.charAt(j)<='z')
				{
					if(p.charAt(j) == s.charAt(i))
						List[i+1][j+1] = List[i][j];
					else List[i+1][j+1] = false;
				}
				else if(p.charAt(j)=='?')
				{
						List[i+1][j+1] = List[i][j];
				}
				else
				{
					List[i+1][j+1] = List[i][j+1] || List[i+1][j];
				}
			}
		}
		return List[ls][lp];
	}
    
}