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 : ABC
Output :
ABC
AB1
A1C
A2
1BC
1B1
2C
3
Note: 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。
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:
GFG
GF1
G1G
G2
1FG
1F1
2G
3
资源:职业杯
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