c++ - 如何在两个字符串中找到公共字符的数量

Question

If string 1 and string 2 are equal sizes, how would we find the num of common characters and store into some int?

For example:

string A = "abcdabc"
string B = "cabzabc"

The common number of characters is 6 out of 7 (count duplicates).

so far I have:

int count = 0;

for(int I = 0; I < A.size(); I++)
{
    if(A[I] == B[I]
    {
       count++;
     }
}

but when I output count its = 0.

EDIT: got it to work guys, something was wrong with my initialized string but now its fine! Thanks.

Answer 1

What you are currently counting is whether those two char arrays have same character on a certain index. This is obviously not what you wish for. Some options: 1. create map and store number of occurrences of each letter, then compare afterwards 2. iterate trough one array, and for each character there, iterate trough complete other array and count occurrences 3. sort these arrays in alphabetical order and you'll see how many times each letter occurs in each of them

EDIT: Let's run trough your code:

I = 0; A[I] = 'a'; B[I] = 'c' //are they the same? count = 0
I = 1; A[I] = 'b'; B[I] = 'a' //are they the same? count = 0
I = 2; A[I] = 'c'; B[I] = 'b' //are they the same? count = 0
I = 3; A[I] = 'd'; B[I] = 'z' //are they the same? count = 0
I = 4; A[I] = 'a'; B[I] = 'a' //are they the same? count = 1
I = 5; A[I] = 'b'; B[I] = 'b' //are they the same? count = 2
I = 6; A[I] = 'c'; B[I] = 'c' //are they the same? count = 3