C 查找所有字符串中存在的公共字符

比较每个字符串，找出所有字符串中常用小写字母的数目。 每个字符串由一个从“a”到“z”的小写字母表示

输入示例： 示例输出： 代码：

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不确定为什么要在这里对前两个字符串进行特殊处理？伪代码中的类似方法如何：

- create a set of characters, name it letters_in_all_strings
- add every lowercase letter to letters_in_all_strings
- for each input string
  - create a set of characters, name it letters_in_this_string
  - add every character in the input string to letters_in_this_string
  - remove all letters from letters_in_all_strings that are not present in letters_in_this_string
- print out the size of letters_in_all_strings

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您可以使用由0和1组成的数组（按字符索引）来实现C中的一组字符，也可以使用glib。或者考虑使用更现代的编程语言？

请考虑一个更具描述性的标题。这个标题可以应用于数千个问题。char var[0][100]；可能不是你想要的。看起来像学校的项目。如果你想在编程方面取得成功，那么就要努力完成这个思维过程。干净利落，切中要害。@BLUEPIXY你能解释var[i][ch-'a']@user3315556 ch-'a'：abcdefghij…->0123456789…，[0..25]记录对应字符的存在情况，并将其替换为字符a-z的索引0-25。如果ch='b'so var[0]['b'-'a']=1'b'-'a'从b到a？@user3315556'b'-'a'为1'在ASCII字符代码中，b'比a'的代码大1。

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

int main() {
    int n;
    scanf("%d",&n);

    char str[n][100];        
    char var[n][26];
    memset(&var[0][0], 0, sizeof(var));

    for(int i=0; i<n; i++) {
        scanf("%99s", str[i]);
        char ch;
        for(int j=0; ch=str[i][j]; ++j){
            if(islower(ch)){
                var[i][ch-'a']=1;//depend on the sequence of character codes
            }
        }
    }

    int x = 0;
    for(int i=0; i<26; ++i){
        int num = 0;
        for(int j=0;j<n;++j)
            if(var[j][i])
                ++num;
        if(num==n)//all string has character of ('a'+i)
            ++x;
    }
    printf("%d\n",x);

    return 0;
}

#include<stdio.h>
#include<string.h>

int main() {
    int n;
    scanf("%d\n",&n);

    char str[n][100];         
    char var[0][100];

    for(int i=0; i<n; i++) { // strings 
        scanf("%99s/n",str[i]);
    }

    for(int i=0;i<100;i++) { // comparison of first 2 strings
        for(int k=0;k<100;k++)
            if(str[0][i]==str[1][k])
                for(int j=0;j<strlen(str[0]);j++) {
                    var[0][j]=str[0][j];          // storing the common letters in a var array
                }
    }

    for(int l=0; l<strlen(str[1]); l++) { // comparing letters in var array with the letters of all other strings                 
        int x;
        if(var[0][l]==str[l+2][l]);
        x=strlen(var[0]);               // counting the common letters
        printf("%d\n",x);
    }

    return 0;
}

- create a set of characters, name it letters_in_all_strings
- add every lowercase letter to letters_in_all_strings
- for each input string
  - create a set of characters, name it letters_in_this_string
  - add every character in the input string to letters_in_this_string
  - remove all letters from letters_in_all_strings that are not present in letters_in_this_string
- print out the size of letters_in_all_strings

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

int main() {
    int n;
    scanf("%d",&n);

    char str[n][100];        
    char var[n][26];
    memset(&var[0][0], 0, sizeof(var));

    for(int i=0; i<n; i++) {
        scanf("%99s", str[i]);
        char ch;
        for(int j=0; ch=str[i][j]; ++j){
            if(islower(ch)){
                var[i][ch-'a']=1;//depend on the sequence of character codes
            }
        }
    }

    int x = 0;
    for(int i=0; i<26; ++i){
        int num = 0;
        for(int j=0;j<n;++j)
            if(var[j][i])
                ++num;
        if(num==n)//all string has character of ('a'+i)
            ++x;
    }
    printf("%d\n",x);

    return 0;
}