Number of combinations of strings

There is a map,key from A to Z, and the corresponding value is from 1 to 26, for example:

A = 1

b = 2

...

Z = 26, for a number string "11", can be a combination of "AA" corresponding to the number of "K" corresponding to the number of the corresponding combination of 2; Similarly, for example, "111" corresponds to the number of combinations is 3, respectively, "AAA", "Ka", "ak".

Given a number array, the corresponding number of combinations is calculated.

Answer: For a string, define f (x) as the number of combinations of the first X characters, then:

f (x) = f (x-2) if the last digit is ' 0 '

= f (x-1) else if (the penultimate number is ' 0 ' | | The last two numbers are greater than "26")

= f (x-1) +f (x-2) Else

intLast2num (Char* Arr,intLen) {Assert (arr! = NULL && len >=2); intnum = (arr[len-2]-'0')*Ten+ (arr[len-1]-'0'); if(Num <= -)        return 2; Else        return 1;}intCountkind (Char* Arr,intLen) {    if(arr = = NULL | | Len <=0)        return 0; if(len = =1)        return 1; if(len = =2)    {        if(arr[len-1] =='0'|| Last2num (arr, len) > -)            return 1; Else             return 2; }    if(arr[len-1] =='0')        returnCountkind (arr, len-2); Else if(arr[len-2] =='0'|| Last2num (arr, len) > -)        returnCountkind (arr, len-1); Else        returnCountkind (arr, len-1) + Countkind (arr, len-2);}

Another way to do this:

The recursion used above, Countkind has repeated calculation, one method is to save the computed in an array, calculate is to check, if counted, directly return the result; Another method is to calculate from 0, the code is as follows:

intLast2num (Char* Arr,intLen) {Assert (arr! = NULL && len >=2); return Ten* (arr[len-2]-'0') + (arr[len-1]-'0');}intCountkind (Char* Arr,intLen) {    if(arr = = NULL | | Len <=0)        return 0; if(len = =1)        return 1; Else if(len = =2)    {        if(arr[len-1] =='0'|| Last2num (arr, len) > -)            return 1; Else            return 2; }    intcnt[len+1]; cnt[0] =0; cnt[1] =1; if(arr[1] =='0'|| Last2num (arr,2) > -) cnt[2] =1; Elsecnt[2] =2;  for(inti =3; i < Len; i++)    {        if(arr[i-1] =='0') Arr[i]= arr[i-2]; Else if(arr[i-2] =='0'|| Last2num (arr,2) > -) Arr[i]= arr[i-1]; ElseArr[i]= arr[i-1] + arr[i-2]; }    returnArr[len];}

A separate calculation of length equal to 2 is necessary in order to solve a case like "70".

Number of combinations of strings