题干
只有一行，是一个字符串，由长度不超过255个字符的小写英文字母组成。
输出格式:
有若干行，每行由两部分组成：一个字母和该字母出现的频率，中间用一个空格分隔，并按频率高低排列，频率相同时则按字母的ASCII码的先后顺序排列。
输入样例:
soon
输出样例:
o 2
n 1
s 1
解题过程
#include <stdio.h>
#include <string.h>
struct CharFreq {
char letter;
int frequency;
};
int compare(const void *a, const void *b) {
struct CharFreq *charFreqA = (struct CharFreq *)a;
struct CharFreq *charFreqB = (struct CharFreq *)b;
if (charFreqA->frequency != charFreqB->frequency) {
return charFreqB->frequency - charFreqA->frequency;
} else {
return charFreqA->letter - charFreqB->letter;
}
}
int main() {
char input[256];
scanf("%s", input);
int freq[26] = {
0};
int len = strlen(input);
for (int i = 0; i < len; ++i) {
if (input[i] >= 'a' && input[i] <= 'z') {
freq[input[i] - 'a']++;
}
}
struct CharFreq charFreqs[26];
for (int i = 0; i < 26; ++i) {
charFreqs[i].letter = 'a' + i;
charFreqs[i].frequency = freq[i];
}
qsort(charFreqs, 26, sizeof(struct CharFreq), compare);
for (int i = 0; i < 26; ++i) {
if (charFreqs[i].frequency > 0) {
printf("%c %d\n", charFreqs[i].letter, charFreqs[i].frequency);
}
}
return 0;
}
