在这里，我们将看到一个与给定字符串的字母数字缩写相关的有趣问题。字符串长度小于10。我们将打印出所有的字母数字缩写。

字母数字缩写是由字符和数字混合形成的。该数字的值是被省略的字符数。可能有任意数量的被省略的子字符串。没有两个子字符串是相邻的。让我们看一下获取这个概念的算法。

算法

printAbbreviation(s, index, max, str) −

begin
   if index is same as max, then
      print str
   end if
   add s[index] at the last of str
   printAbbreviation(s, index + 1, max, str)
   delete last character from str
   count := 1
   if str is not empty, then
      if the last character of str is a digit, then
         add last digit with the count value
         delete last character from str
      end if
   end if
   add count after the str
   printAbbreveation(s, index + 1, max, str)
end

示例

#include <iostream>
using namespace std;
void printAbbreviation(const string& s, int index, int max_index, string str) {
   if (index == max_index) { //if string has ended
      cout << str << endl;
      return;
   }
   str.push_back(s[index]); // push the current character to result
   printAbbreviation(s, index + 1, max_index, str); //print from next index
   str.pop_back(); //remove last character
   int count = 1;
   if (!str.empty()) {
      if (isdigit(str.back())) { //if the last one is digit, then
         count += (int)(str.back() - '0'); //count the integer value of that digit
         str.pop_back(); //remove last character
      }
   }
   char to_char = (char)(count + '0'); //make count to character
   str.push_back(to_char);
   printAbbreviation(s, index + 1, max_index, str); //do for next index
}
void printCombination(string str) {
   if (!str.length()) //if the string is empty
      return;
   string str_res;
   printAbbreviation(str, 0, str.length(), str_res);
}
int main() {
   string str = "HELLO";
   printCombination(str);
}

输出

HELLO
HELL1
HEL1O
HEL2
HE1LO
HE1L1
HE2O
HE3
H1LLO
H1LL1
H1L1O
H1L2
H2LO
H2L1
H3O
H4
1ELLO
1ELL1
1EL1O
1EL2
1E1LO
1E1L1
1E2O
1E3
2LLO
2LL1
2L1O
2L2
3LO
3L1
4O
5