计算字符串中每个字母的出现次数

Question

如何计算字符串中每个字母(忽略大小写)中c的出现次数？因此它会打印出来letter: # number of occurences,我有代码来计算一个字母的出现次数,但是如何计算字符串中每个字母的出现次数呢？

{
    char
    int count = 0;
    int i;

    //int length = strlen(string);

    for (i = 0; i < 20; i++)
    {
        if (string[i] == ch)
        {
            count++;
        }
    }

    return count;
}

输出:

a : 1
b : 0
c : 2
etc...

Answer 1

假设您有一个char8位的系统,并且您尝试计算的所有字符都使用非负数进行编码.在这种情况下,您可以写:

const char *str = "The quick brown fox jumped over the lazy dog.";

int counts[256] = { 0 };

int i;
size_t len = strlen(str);

for (i = 0; i < len; i++) {
    counts[(int)(str[i])]++;
}

for (i = 0; i < 256; i++) {
    printf("The %d. character has %d occurrences.\n", i, counts[i]);
}

请注意,这将计算字符串中的所有字符.如果你100%绝对肯定你的字符串里面只有字母(没有数字,没有空格,没有标点符号),那么1.要求"不区分大小写"开始有意义,2.你可以减少条目数到英文字母中的字符数(即26),你可以这样写:

#include <ctype.h>
#include <string.h>
#include <stdlib.h>

const char *str = "TheQuickBrownFoxJumpedOverTheLazyDog";

int counts[26] = { 0 };

int i;
size_t len = strlen(str);

for (i = 0; i < len; i++) {
    // Just in order that we don't shout ourselves in the foot
    char c = str[i];
    if (!isalpha(c)) continue;

    counts[(int)(tolower(c) - 'a')]++;
}

for (i = 0; i < 26; i++) {
    printf("'%c' has %2d occurrences.\n", i + 'a', counts[i]);
}