Question

我有两个字符串a，b，长度分别为l1，l2。我想返回给定两个字符串的常用字母数。例如，a='ABC'和b='CDE'。我的算法应返回1，因为只有'C'是两个字符串中的公共字母，但它会返回26。有人可以解释一下原因吗？这是我的算法：

for(i=0; i < l1; i++)
{
    for(j=0; j < l2; j++)
    {
        if(a[i] == b[j])
        {
            found++;
        }
    }
}

Answer 1

您的算法会失败您的标准。如果一个字符串是'CCC'，很容易看出会发生什么。

这样做：

int letters[26]={0};
int both[26]={0};

for(int i=0; i<l1; i++)
   letters[a[i]-'A']=1;

for(int i=0; i<l2; i++)
    if(letters[b[i]-'A'])
        both[b[i]-'A']=1;

int found=0;

for(int i=0; i<26; i++)
    found+=both[i];

Answer 2

// With this code I preferred to convert all char as an ascii code.
// Because it is more simpler to compare int value than with `2` for statement 
// it has compared if chars are similar, increase the number of `sayi`, then 
// print the equal char of these code

int main(){
    char a[]="ABC";
    char b[]="CDE";
    int sizeofa=sizeof(a)-1;
    int sizeofb=sizeof(b)-1;
    int sayi=0;
    int asciia;
    int asciib;

    for (int i=0; i<sizeofa;i++)
    {
        for (int j=0; j<sizeofb; j++)
        {
            asciia=a[i];
            asciib=b[j];

            if ( asciia == asciib)
                sayi++;
        }

    }
    printf("%d",sayi);
}

Answer 3

示例不完整，因此很难说出为什么会得到26。可能found尚未初始化或您的打印输出不正确。

算法本身不正确，因为它会过度计算常见字符。例如

char *a = "A";
char *b = "AAAA";

第二个循环将计算'A'个字符4次。

for(j=0;j<l2;j++)
{
    if(a[i]==b[j])

你必须避免过度计算，例如看看：

size_t count_common_chars(const char *s1, const char *s2) {
    size_t counter = 0;   // our occurrence counter
    int index[256] = {0}; // clear the index table

    while(*s1){           // for every character in the string s1
        index[*s1]++;     // use string s1 characters as index to index table  s1 = "aaa" would yield index['a'] == 3 
        s1++;             // advance to next char
    }

    // For every character in the string s2 check if character existed in s1.
    // If yes increase the counter. 
    // 
    // If a particular character occurs 'n' times in string s1
    // then the number of common characters in string s2 cannot be greater than 'n'. 
    // That is why we have to subtract the number of occurrences of a given character 'X' in string s2
    // from the reference index table which counted the occurrence of the same character in string s1.
    //
    // This is a generic approach, we would do subtraction whether or not there is a match.
    // At the end the index table contains a count of a differences between s1 and s2.
    // Positive number at the specific index means that there was more such characters in string s1 
    // Negative number at the specific index means that there was more such characters in string s2
    // 0 means that number of particular character was the same or there was no such character in the strings s1 and s2

    while(*s2){            // for every character in the string s2
        if ( (index[*s2]--)  >  0)  
            counter++;
        s2++;
    }            
    return counter;
}

Answer 4

 It's return 1 
 char a[3]={'a','b','c'};
        char b[3]={'c','d','e'};
    int found=0,l1=3,l2=3;
       for(int i=0;i<l1;i++)
        {
            for(int j=0;j<l2;j++)
            {
                if(a[i]==b[j])
                {
                    found++;
                }
            }
        }

用于搜索字符串匹配的算法

4 个答案: