问题：将一连串的字符转化成用数字表示的数字串 ，如：
3234567 ----- 323-4567
888-GINO ----- 888-1010
3-10-10-10 ----- 310-1010
-8-2-3-5-7-2-9- -- 823-5729
该问题的关键点在于去 ‘-’ 和 将对应的字符映射到数字，见代码：
1 /************************************************************************/ 2 * Author: bakari 3 * Date :2012/6/5 4 * StandardPhoneNum 5 /************************************************************************/ 6 #include <iostream> 7 #include <string.h> 8 #include <algorithm> 9 using std::cout; 10 using std::cin; 11 using std::endl; 12 13 const int STRLEN = 9; 14 const int MAXSTRLEN = 50; 15 const int MAXROW = 10000; 16 17 char map[] = "22233344455566677778889999"; //存储映射表 18 char TranseStr[MAXROW][STRLEN]; 19 //char str[MAXSTRLEN]; 20 21 int cmp (const void * elem1, const void * elem2){ //快排比较函数 22 return strcmp((char *)elem1, (char *)elem2); 23 24 } 25 void StandardTranse(const char *str, int n){ 26 27 for(size_t i = 0, j = 0; j < STRLEN - 1; ){ 28 if(str[i] != '-'){ 29 if(str[i] <= 'Z' && str[i] >= 'A') 30 TranseStr[n][j] = map[str[i] - 'A']; 31 if(str[i] <= '9' && str[i] >= '0') 32 TranseStr[n][j] = str[i]; 33 ++ i; 34 ++ j; 35 } 36 else{ 37 ++ i; 38 } 39 if(3 == j){ 40 TranseStr[n][j++] = '-'; 41 } 42 } 43 } 44 45 void FindSameNum(int n){ 46 int ix = 0; 47 int jx; 48 bool noduplicate = true; 49 while(ix < n){ 50 jx = ix; 51 ++ ix; 52 while(ix < n && strcmp(TranseStr[ix],TranseStr[jx]) == 0) 53 ++ix; 54 if (ix - jx > 1){ 55 cout << TranseStr[jx] << " " << ix - jx <<endl; 56 noduplicate = false; 57 } 58 } 59 if (noduplicate) 60 cout << "No duplicate." << endl; 61 } 62 63 int main() 64 { 65 int n; 66 char str[MAXSTRLEN]; 67 cin >> n; 68 for(int i = 0; i != n; ++i){ 69 cin >> str; 70 StandardTranse(i); 71 } 72 cout << "输出" <<endl; 73 for(int i = 0; i != n; ++i) 74 cout << TranseStr[i] << endl; 75 76 qsort(TranseStr,n,STRLEN,cmp); 77 FindSameNum(n); 78 return 0; 79 }
标准化函数StandardTranse()的另一个解法
1 void StandardTranse(const char *str, int n){ 2 int j ,k; 3 j = k = -1; 4 while(k < STRLEN - 1){ 5 j ++; 6 if(str[j] == '-') //先做判断 7 continue; 8 k ++; 9 if(3 == k) 10 TranseStr[n][k ++] = '-'; 11 if(str[j] >= 'A' && str[j] <= 'Z'){ 12 TranseStr[n][k] = map[str[j] - 'A']; 13 continue; 14 } 15 TranseStr[n][k] = str[j]; 16 } 17 TranseStr[n][k] = '\0'; 18 }
