给定文本和通配符模式，实现通配符模式匹配算法，以查找通配符模式是否与文本匹配。匹配应涵盖整个文本(而不是部分文本)。
通配符模式可以包含字符“?”、“*”和“+”。
‘?’ – matches any single character
‘*’ – Matches any sequence of characters
(including the empty sequence)
'+' – Matches previous single character
of pattern 例子：
Input :Text = "baaabaaa",
Pattern = “****+ba*****a+", output : true
Pattern = "baaa?ab", output : false
Pattern = "ba*a?", output : true
Pattern = "+a*ab", output : false
Input : Text = "aab"
Pattern = "*+" output : false
Pattern = "*+b" output : true
情况一：字符为’*’
这里出现两种情况：我们可以忽略 ‘*’字符并移动到模式中的下一个字符。 ‘*’字符与文本中的一个或多个字符匹配。在这里，我们将移至字符串的下一个字符。
情况 2：字符是 ‘?’我们可以忽略文本中的当前字符并移动到模式和文本中的下一个字符。
情况 3：字符为 ‘+’
这里出现两种情况：我们将文本的当前字符与模式的前一个字符匹配。如果没有前面的字符意味着 ‘+’ 是模式的第一个字符，那么我们将结果打印为“文本不匹配”。如果前一个字符是“+”、“?”或 ‘*’ 我们用他们使用的最后一个字符替换它。
案例4：字符不是字符
如果在文本当前字符与图形当前字符相匹配，我们移动到图形和文本下一个字符。如果它们不匹配，则通配符模式和文本不匹配。
“*”、“?”的过程类似于两个字符的通配符模式匹配：
Here we use a dp table that will contain
two fields
Struct DP
{
// value is true if match possible
// for current indexes, else false.
bool value;
// Stores the character used
// by the symbol that we used
// later for symbol '+'
char ch;
}
下面是上述想法的 C++ 实现。
// C++ program to implement wildcard
// pattern matching algorithm
#include
using namespace std;
struct DP {
bool value;
char ch;
};
// Function that matches input str with
// given wildcard pattern
bool strmatch(string str, string pattern,
int n, int m)
{
// empty pattern can only match with
// empty string
if (m == 0)
return (n == 0);
// If first character of pattern is '+'
if (pattern[0] == '+')
return false;
// lookup table for storing results of
// subproblems
struct DP lookup[m + 1][n + 1];
// initialize lookup table to false
for (int i = 0; i <= m; i++)
for (int j = 0; j <= n; j++) {
lookup[i][j].value = false;
lookup[i][j].ch = ' ';
}
// empty pattern can match with
// empty string
lookup[0][0].value = true;
// Only '*' can match with empty string
for (int j = 1; j <= n; j++)
if (pattern[j - 1] == '*')
lookup[j][0].value =
lookup[j - 1][0].value;
// fill the table in bottom-up fashion
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
// Two cases if we see a '*'
// a) We ignore ‘*’ character and move
// to next character in the pattern,
// i.e., ‘*’ indicates an empty sequence.
// b) '*' character matches with ith
// character in input
if (pattern[i - 1] == '*') {
lookup[i][j].value =
lookup[i][j - 1].value ||
lookup[i - 1][j].value;
lookup[i][j].ch = str[j - 1];
}
// Current characters are considered as
// matching in two cases
// (a) current character of pattern is '?'
else if (pattern[i - 1] == '?') {
lookup[i][j].value =
lookup[i - 1][j - 1].value;
lookup[i][j].ch = str[j - 1];
}
// (b) characters actually match
else if (str[j - 1] == pattern[i - 1])
lookup[i][j].value =
lookup[i - 1][j - 1].value;
// Current character match
else if (pattern[i - 1] == '+')
// case 1: if previous character is
// not symbol
if (pattern[i - 2] != '+' ||
pattern[i - 2] != '*' ||
pattern[i - 2] != '?')
if (pattern[i - 2] == str[j - 1]) {
lookup[i][j].value =
lookup[i - 1][j - 1].value;
lookup[i][j].ch = str[j - 1];
}
// case 2 : if previous character
// is symbol (+, *, ? ) then we
// compare current text character
// with the character that is used by
// the symbol at that point. we
// access it by lookup[i-1][j-1]
else if (str[j-1] == lookup[i-1][j-1].ch) {
lookup[i][j].value =
lookup[i - 1][j - 1].value;
lookup[i][j].ch =
lookup[i - 1][j - 1].ch;
}
// If characters don't match
else
lookup[i][j].value = false;
}
}
return lookup[m][n].value;
}
// Driver code
int main()
{
string str = "baaabaaa";
string pattern = "*****+ba***+";
if (strmatch(str, pattern, str.length(),
pattern.length()))
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
输出：
Yes
时间复杂度：O(n*m)
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