c - 带指针的二维数组中最常见的字符

Question

我想找到5个字符串中最常见的字符，我想到了这一点，我已经为特定的字符完成了它，在我的场合，但是我如何为每个字符做它，它只是我的代码的一部分

Ch 是一个指针!!!

for(i = 0; i < five; i++){
    ch = strchr(cities[i], 's'); 

    if (ch != NULL){ //if this is true 
        ch1++;
        printf("exists in %s\n", cities[i]); //print
    }
    else
    {
        printf ("doesnt exist in %s\n", cities[i]); //print
    }
}
printf("The character 's' appears in %d cities\n", ch1);

Answer 1

您必须以某种方式将每个字符出现的次数存储在某处。

给出的提示是保留一个变量来存储字符的最大出现次数和字符本身。没有其他的。对于循环，您需要考虑 a 中的所有字符至z .

代码将类似于:-(仅在您觉得自己已经尝试了足够多的情况下才查看)。

为了让您了解此代码的作用 - 它计算出现在最大城市数中的字符。如果存在平局，它首先选择较小的字母表(按字典顺序)。

for( char c = 'a'; c <= 'z';c++){
    int ch1 = 0, mxch1 = 0;
    char mch='?';
    for(i = 0; i < numOfCities; i++){
      ch=strchr(cities[i],c); 
        if( ch ){
           printf("Exist in %s\n",cities[i]);
           ch1++;
        }
        else
           printf("Doesn't Exist in %s\n",cities[i]);
    }
    if( ch1 > mch1) { 
      mch1 = ch1; 
      mch = c; 
  }
    printf("The character %c appears in %d cities\n",c,ch1);
}
printf("Character which occured in max number of cities is %c %d",mch,mch1);

最常见字符的代码会有点不同。

int charMap[26]={0};
for(size_t i = 0; i < numOfCities; i++){
  for(size_t j = 0; cities[i][j]; j++)
      charMap[cities[i][j]-'a']++;
}
int mx = 0;
mxi = 0;
for(size_t i = 0; i < sizeof(charMap)/sizeof(charMap[0]); i++){
  if(charMap[i]>mx){
      mx  = charMap[i];
      mxi = i;
  }

printf("Character which occured max number of times is %c %d", (char)(mxi+'a'),mx);