可用于在C / C ++中拨号的字符串的所有组合吗？
对于给定的号码，显示或打印所有可能的字符串组合，可以在以下说明的帮助下实现在电话中拨打给定号码的所有字符串组合。
在给定的电话中，我们可以拨打，
2实现A或B或C，
3实现D或E或F，
………………。
8实现T或U或V，
9实现W或X或Y或Z，
1仅实施1
0实现0。
例如，如果给定的电话号码是89，则程序应打印
TW，TX，TY，TZ，UW，UX，UY，UZ，VW，VX，VY，VZ
#include <stdio.h>
#include <string.h>
// TableHash[i] stores all characters that correspond to digit i in phone const char TableHash[10][5] =
{"", "", "WXYZ", "TUV", "PQRS", "MNO", "GHI", "GHI", "DEF", "ABC"};
// A recursive function to display or print all possible words that can be obtained by implementing input number1[] of size n1.
The output words are one by one stored in output1[] void UtilWordsPrint(int number1[], int curr_digit1, char output1[], int n1) {
// In the Base case, if current output word is prepared int i; if (curr_digit1 == n1) {
printf("%s ",
output1); return ;
}
// We try all 3 possible characters for current digit in number1[]
// and recur for remaining digits
for (i=0; i<strlen(TableHash[number1[curr_digit1]]); i++) {
output1[curr_digit1] =
TableHash[number1[curr_digit1]][i];
UtilWordsPrint(number1, curr_digit1+1, output1, n1);
if (number1[curr_digit1] == 0 || number1[curr_digit1] == 1)
return;
}
}
// A wrapper over UtilWordsPrint(). It is able to create an output1 array and calls UtilWordsPrint() void printWords(int number1[], int n1) {
char result1[n1+1];
result1[n1] ='\0';
UtilWordsPrint(number1, 0, result1, n1);
}
//Driver program int main(void) {
int number1[] = {2, 3, 4};
int n1 = sizeof(number1)/sizeof(number1[0]);
printWords(number1, n1); return 0;
}输出结果
WTP WTQ WTR WTS WUP WUQ WUR WUS WVP WVQ WVR WVS XTP XTQ XTR XTS XUP XUQ XUR XUS XVP XVQ XVR XVS YTP YTQ YTR YTS YUP YUQ YUR YUS YVP YVQ YVR YVS ZTP ZTQ ZTR ZTS ZUP ZUQ ZUR ZUS ZVP ZVQ ZVR ZVS
时间复杂度：上述代码beO(4n)的时间复杂度，其中n被视为输入数字中的位数。