題目一

請實現一個函數用來匹配包括‘.‘和‘*‘的正則表達式。模式中的字符‘.‘表示任意一個字符，而‘*‘表示它前面的字符可以出現任意次（包含0次）。 在本題中，匹配是指字符串的所有字符匹配整個模式。例如，字符串"aaa"與模式"a.a"和"ab*ac*a"匹配，但是與"aa.a"和"ab*a"均不匹配

思路

用動態規劃來解決

規定

若，則str[0~i-1]和pattern[0~j-1]匹配

1、當p[j-1] != * 時

2、當p[j-1] = * 時

　　記p[j-2]=X,分為以下兩種情況

　　(a) "X*"重復了0次：等式右邊第一項說明*重復零次，則f[i][j]與 f[i][j-2]真假相同，亦或是p[j-2]是萬能字符

　　(b)"X*"重復了大於等於1次：

　　　　等式右邊第一項說明要麽是萬能字符，

　　　　第二項表示f[i][j]與f[i-1][j]真假相同，其實就是回退到重復0次的情況

　　　　第三項保證復制的正確性

　　　　如 s=bcaaa p=bca* f[5][4] -> f[4][4] -> f[3][4] -> f[2][4]

　　　　此時 s = bc p = bca*，可以匹配

算法流程

1、初始化二維布爾數組，註意列不一定為0

2、按以上規則自下網上計算出f[i][j]

計算過程

class Solution {
public:
    bool match(char* str, char* pattern)
    {
        int m = strlen(str);
        int n = strlen(pattern);
        vector<vector<bool>> f(m+1, vector<bool>(n+1, false));
        f[0][0] = true
 
;
        
        for (int i = 1; i <= m; i++)
        {
            f[i][0] = false;
        }
        for (int j = 1; j <= n; j++)
        {
            f[0][j] = ((j > 1) && (pattern[j-1] == ‘*‘) && (f[0][j-2]));
        }
        
        for (int i = 1; i <= m; i++){
            for (int j = 1; j <= n; j++){
                if (pattern[j-1] != ‘*‘)
                    f[i][j] = f[i - 1][j - 1] && (str[i - 1] == pattern[j - 1] || ‘.‘ == pattern[j - 1]);
                else
                    f[i][j] = f[i][j-2] || (pattern[j-2] == ‘.‘) || (str[i-1] == pattern[j-2]) && f[i-1][j];
            }
        }
        return f[m][n];
    }
};

題目二

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).

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

class Solution {
public:
    bool isMatch(string s, string p) {
        int n = (int)s.length();
        int m = (int)p.length();
        
        vector<vector<bool>> dp(n + 1, vector<bool>(m + 1));
        
        dp[0][0] = true;
        for (int i = 1; i <= m && p[i - 1] == ‘*‘; i++)
            dp[0][i] = true;

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (p[j - 1] == ‘*‘)
                    dp[i][j] =  dp[i - 1][j] || dp[i][j - 1];
                else
                    dp[i][j] = (p[j - 1] == ‘?‘ || s[i - 1] == p[j - 1]) && dp[i - 1][j - 1];
            }
        }
        
        return dp[n][m];
    }
};

【劍指offer】19、正則表達式匹配 && 【Leetcode】44、Wildcard Matching