C语言编程-9_4 字符统计

Posted 2022-01-02 ly570

输入一个字符串（其长度不超过81），分别统计其中26个英文字母出现的次数（不区分大、小写字母），并按字母出现次数从高到低排序，若次数相同，按字母顺序排列。字母输出格式举例，例如：A-3，表示字母A出现3次，C-0表示字母C没有出现。

       输入：

       第一行为输入，占一行，输入的字符串可能含有空格

       输出：

       第二行为输出，占一行。按照字母输出格式从高到低输出，各字母输出之间用一个空格字符分隔。

       样例：

       123abcAABXxwvUu+

       A-3 B-2 U-2 X-2 C-1 V-1 W-1 D-0 E-0 F-0 G-0 H-0 I-0 J-0 K-0 L-0 M-0 N-0 O-0   P-0 Q-0 R-0 S-0 T-0 Y-0 Z-0 

//插入排序法
#include<stdio.h>
#include<string.h>


int main()

char str[81] = "123abcAABXxwvUu+";
gets(str);
int letter_stat[26] = 0;
char letter[26] = ‘ ‘ ;
for (int i = 0; i < 26; i++)

letter[i] = i + ‘A‘;

int i = 0;
while (str[i])

int k = 0;
if (str[i] >= ‘A‘&&str[i] <= ‘Z‘)

k = str[i] - ‘A‘;
letter_stat[k]++;

else if (str[i] >= ‘a‘&&str[i] <= ‘z‘)

k = str[i] - ‘a‘;
letter_stat[k]++;

i++;


for (int i = 0; i < 26; i++)

int k = letter_stat[i];
int temp = letter[i];
int j = i - 1;
while (j >= 0 && letter_stat[j] < k)

letter_stat[j + 1] = letter_stat[j];
letter[j + 1] = letter[j];
j--;

letter_stat[j + 1] = k;
letter[j + 1] = temp;

for (int i = 0; i < 25; i++)

printf("%c-%d ", letter[i],letter_stat[i]);

printf("%c-%d", letter[25], letter_stat[25]);
return 0;

//冒泡排序法
#include<stdio.h>
#include<string.h>


int main()

char str[81] = "123abcAABXxwvUu+";
gets(str);
int letter_stat[26] = 0;
char letter[26] = ‘ ‘ ;
for (int i = 0; i < 26; i++)

letter[i] = i + ‘A‘;

int i = 0;
while (str[i])

int k = 0;
if (str[i] >= ‘A‘&&str[i] <= ‘Z‘)

k = str[i] - ‘A‘;
letter_stat[k]++;

else if (str[i] >= ‘a‘&&str[i] <= ‘z‘)

k = str[i] - ‘a‘;
letter_stat[k]++;

i++;

for (int i = 0; i < 25; i++)
for (int j = 0; j < 25 - i; j++)


if (letter_stat[j] < letter_stat[j + 1])

int temp = letter_stat[j + 1];
letter_stat[j + 1] = letter_stat[j];
letter_stat[j] = temp;

temp = letter[j + 1];
letter[j + 1] = letter[j];
letter[j] = temp;





for (int i = 0; i < 25; i++)

printf("%c-%d ", letter[i],letter_stat[i]);

printf("%c-%d", letter[25], letter_stat[25]);
return 0;

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