想知道为什么用switch case没过oj，用if语句过了？？

Question

//if()版本
#include<stdio.h>
int main()
{
   char str[5000]={'\0'};
   gets(str);
   int sum=0;
   for(int i=0;str[i];i++)
   {
       int numble=str[i];
       if(str[i] == 'a'||str[i] == 'd'||str[i] == 'g'||str[i] == 'j'||str[i] == 'm'||str[i] == 'p'||str[i] == 't'||str[i] == 'w'||str[i] == ' ')
            sum = sum + 1;
       else if(str[i] == 'b'||str[i] == 'e'||str[i] == 'h'||str[i] == 'k'||str[i] == 'n'||str[i] == 'q'||str[i] == 'u'||str[i] == 'x')
            sum =sum + 2;
       else if(str[i] == 'c'||str[i] == 'f'||str[i] == 'i'||str[i] == 'l'||str[i] == 'o'||str[i] == 'r'||str[i] == 'v'||str[i] == 'y')
            sum =sum + 3;
       else if(str[i] ==  's'||str[i] == 'z')
            sum = sum + 4;

   }
   printf("%d",sum);
}
//switch case版本

#include<stdio.h>
int character_times(int character);
int main()
{
   char str[5000]={'\0'};
   gets(str);
   int sum=0;
   for(int i=0;str[i];i++)
   {
       int numble=str[i];
       sum+=character_times(numble);

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

int character_times(int character)//次数
{
     int times=0;
     switch(character)
        {
            case'a':
            case'd':
            case'g':
            case'j':
            case'm':
            case'p':
            case't':
            case'w':
                times=1;
                break;
            case'b':
            case'e':
            case'h':
            case'k':
            case'n':
            case'q':
            case'u':
            case'x':
                times=2;
                break;
            case'c':
            case'f':
            case'i':
            case'l':
            case'o':
            case'r':
            case'v':
            case'y':
                times=3;
                break;
            case's':
            case'z':
                times=3;
                break;
            case ' ':
                times=1;

        }
        int ret=times;
        return ret;
}

Answer 1

第 78行是 times = 4啊