C语言 获取字符串中字符出现次数

1

#include <stdio.h>
void main()
{
    char str[80];
    int tj[26] = {0};
    int i;
    gets(str);
    for (i = 0; str[i] != '\0'; i++)
        if (str[i] >= 'a' && str[i] <= 'z')
            tj[str[i] - 'a']++;
    for (i = 0; i < 26; i++)
        if (tj[i] != 0)
            printf("%c=%d  ", 'a' + i, tj[i]);
}

2

#include <stdio.h>

struct word
{
    char w;
    int c;
};

int main()
{

    char str[100] = "abc^db";
    // scanf("%s", str);
    struct word arr[26];
    int i = 0, arrlen = 0;

    while (str[i])
    {
        if (!(str[i] >= 'a' && str[i] <= 'z'))
        {
            i++;
            continue;
        }
        int index = -1;
        int j = 0;
        for (j = 0; j < arrlen; j++)
        {
            if (str[i] == arr[j].w)
            {
                index = j;
                break;
            }
        }

        if (index == -1)
        {
            arr[arrlen].w = str[i];
            arr[arrlen].c = 1;
            arrlen++;
        }
        else
            arr[index].c += 1;

        i++;
    }

    //输出
    for (int x = 0; x < arrlen; x++)
    {
        printf("%c = %d ", arr[x].w, arr[x].c);
    }

    return 0;
}