字符串的字母数字缩写 _字符串

给定一个长度小于10的字符串, 我们需要打印该字符串的所有字母数字缩写。
【字符串的字母数字缩写】字母数字缩写形式为字符与数字混合的形式, 该数字等于所选子字符串的跳过字符数。因此, 每当将跳过字符的子字符串, 你必须将其替换为表示子字符串中字符数的数字。字符串中可以有任意多个跳过的子字符串。没有两个子字符串彼此相邻。因此, 结果中没有两位数字相邻。有关更清晰的主意, 请参见示例。
例子：

Input : ANKS Output :ANKS (nothing is replaced)ANK1 (S is replaced) AN1S (K is replaced)AN2(KS is replaced)A1KS (N is replaced)A1K1 (N and S are replaced)A2S (NK is replaced)A3 (NKS is replaced)1NKS (A is replaced)1NK1 (A and S are replaced)1N1S (A and N is replaced)1N2 (A and KS are replaced)2KS (AN is replaced)2K1 (AN and S is replaced)3S (ANK is replaced)4 (ANKS is replaced)Input : ABCOutput : ABCAB1 A1C A2 1BC 1B1 2C 3Note: 11C is not valid because no two digits should be adjacent, 2C is the correct one because AB is a substring, not A and B individually

资源：Google面试问题
推荐：请尝试以下方法{IDE}首先, 在继续解决方案之前。想法是从空字符串开始。在每一步中, 我们都有两个选择。

按原样考虑字符。
添加字符进行计数。如果没有计数, 请使用1。

文章图片
你可以看到每个字符如何以字符或数字的形式累加到结果中。这进一步在末尾引起2 ^ n缩写, 其中n是字符串的长度。

// C++ program to print all Alpha-Numeric Abbreviations // of a String #include < bits/stdc++.h> using namespace std; // Recursive function to print the valid combinations // s is string, st is resultant string void printCompRec( const string& s, int index, int max_index, string st) { // if the end of the string is reached if (index == max_index) { cout < < st < < "\n" ; return ; }// push the current character to result st.push_back(s[index]); // recur for the next [Using Char] printCompRec(s, index + 1, max_index, st); // remove the character from result st.pop_back(); // set count of digits to 1 int count = 1; // addition the adjacent digits if (!st.empty()) {if ( isdigit (st.back())) {// get the digit and increase the count count += ( int )(st.back() - '0' ); // remove the adjacent digit st.pop_back(); } }// change count to a character char to_print = ( char )(count + '0' ); // add the character to result st.push_back(to_print); // recur for this again [Using Count] printCompRec(s, index + 1, max_index, st); }// Wrapper function void printComb(std::string s) { // if the string is empty if (!s.length()) return ; // Stores result strings one one by one string st; printCompRec(s, 0, s.length(), st); }// driver function int main() { string str = "GFG" ; printComb(str); return 0; }

输出如下：