给定文本和通配符模式，请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。匹配项应覆盖整个文本(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1：字符为“ *”
这里出现两种情况：我们可以忽略’*’字符并移至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，而移至图案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进行匹配。如果没有先前的字符，则意味着’+’是模式的第一个字符，因此我们将结果打印为“文本不匹配”。如果前一个字符是“ +”，“?”或“ *”，我们将其替换为他们使用的最后一个字符。
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。
“ *”，“?”的过程与通配符模式匹配两个字符相似：
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;
}
下面的C++实现了上面的想法。
// 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)