【leetcode】44. Wildcard Matching

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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

与leetcode上面的第10题正则匹配表达式不同，这道题的模式p的第一个字符是有可能出现*号的，而不像正则表达式那样*号只出现在.号后面。采用动态规划思想对该题进行解答。

具体代码如下：

class Solution {
    public boolean isMatch(String s, String p) {
        boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
        dp[s.length()][p.length()] = true;
        //考虑到p模式第一个字符为*
        for(int i = p.length() - 1;i >= 0;i--) {
            if(p.charAt(i) == '*') {
                dp[s.length()][i] = true;
                //break;
            } else {
                break;
                //dp[s.length()][i] = true;
            }
        }
        
        //dp[i][j]代表s从i开始截取，p从j开始截取
        for(int i = s.length() - 1;i >= 0;i--) {
            for(int j = p.length() - 1;j >= 0;j--) {
                if(s.charAt(i) == p.charAt(j) || p.charAt(j) == '?') {
                    dp[i][j] = dp[i + 1][j + 1];
                } else if(p.charAt(j) == '*') {
                    dp[i][j] = dp[i+1][j] || dp[i][j+1];
                } else {
                    dp[i][j] = false;
                }
            }
        }
        
        return dp[0][0];
    }
}

10. Regular Expression Matching 

leetcode上面大佬的代码，对比着看

class Solution {
    public boolean isMatch(String str, String pattern) {
        int s = 0, p = 0, match = 0, starIdx = -1;            
        while (s < str.length()){
            // advancing both pointers
            if (p < pattern.length()  && (pattern.charAt(p) == '?' || str.charAt(s) == pattern.charAt(p))){
                s++;
                p++;
            }
            // * found, only advancing pattern pointer
            else if (p < pattern.length() && pattern.charAt(p) == '*'){
                starIdx = p;
                match = s;
                p++;
            }
           // last pattern pointer was *, advancing string pointer
            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();
    }
}