【C】关于字符串与字符串函数de一些小练习

关于字符串与字符串函数de小练习

---

1.字符串中的最大数

你需要找出十个数中最大的哪一个，但不幸的是因为一些故障，一些小写字母随机的插入了这是个数字。请忽视这十个字符串中无意义的小写字母，输出这十个数字中最大的那一个，以及它来自于哪一个字符串。

输入：

"a3a2dsa3f4fsa5dgf";
"b2345fsa5dgf";
"2344h25ihiooy2";
"a3a234g5dgf";
"a3a2ds324fdgrhdgf";
"a3fghdfgsddgf";
"sad87f89q374";
"drghe478502";
"2re9fu02987r";
"32wtewr939rgu";

输出：

max number is 8789374
comes from sad87f89q374

题解：使用strpbrk()函数，可以查找一个字符串当中符合一个特定字符集合的字符，并返回一个指向它的指针。借由这种方式找出字符串中的数字，合成一个整数，并找出最大数即可。

#include 
#include 

int get(char const *string);
int find(char **,int *);

int main() {
    char *p[10];
    p[0] = "a3a2dsa3f4fsa5dgf";
    p[1] = "b2345fsa5dgf";
    p[2] = "2344h25ihiooy2";
    p[3] = "a3a234g5dgf";
    p[4] = "a3a2ds324fdgrhdgf";
    p[5] = "a3fghdfgsddgf";
    p[6] = "sad87f89q374";
    p[7] = "drghe478502";
    p[8] = "2re9fu02987r";
    p[9] = "32wtewr939rgu";
    int maxlabel = 0;
    int *plabel = &maxlabel;
    int max = find(p,plabel);

    printf("max number is %d\ncomes from %s\n",max,p[maxlabel]);

    return 0;
}

int find(char **p,int *plabel){
    int num[10],max=0;
    *plabel=0;
    for(int i = 0;i<10;i++){
        num[i] = get(p[i]);
        *plabel = (num[i]>max) ? i : *plabel ;
        max = (num[i]>max) ? num[i] : max ;
    }
    return max;
}

int get(char const *string){//输入字符串，输出转换后的数字

    char *p = strpbrk(string,"0123456789");//指向了第一个数字
    int num = 0 ;
    while(1){
        num *= 10;
        num += ((int)*p-(int)'0');
        string = p + 1;
        if(*string == '\0'){break;}
        p = strpbrk(string,"0123456789");
        if(p == NULL){break;}
    }
    return num;
}

---

2.摘抄好句

阅读一段仅由空格逗号，句号，大小写字母组成的英语文本，并从中找出最长的一句话，将其打印出来。

输入：

"This mentality counts in almost every stage of our life.First, when a student keeps learning new skills, he or she always asks for more and desires to explore the unknown world. Therefore, there is an excellent chance that he or she stands out in peer groups. Second, in the workplace,office workers with such attitude generally finish his or her tasks in a higher quality, and they are more likely to climb up the ladder more quickly than their colleagues who content themselves with the skills they already have.Besides, in our daily life, people who keep learning new skills are more positive, and everyone around them must be fond of making friends with them."

输出：

Longest sentence is
Second, in the workplace,office workers with such attitude generally finish his or her tasks in a higher quality, and they are more likely to climb up the ladder more quickly than their colleagues who content themselves with the skills they already have.

题解：使用strtok()函数可以分割字符串，这里只需要分割句号结尾的就是一个句子了。strlen()函数可以获取字符串长度，借此就可以找出“最长的句子”了。

#include 
#include 
char *longsentence(char * string);

int main(void){
char string[] = "This mentality counts in almost every stage of our life.First,"
" when a student keeps learning new skills, he or she always asks for more and de"
"sires to explore the unknown world. Therefore, there is an excellent chance that he "
"or she stands out in peer groups. Second, in the workplace,office workers with such a"
"ttitude generally finish his or her tasks in a higher quality, and they are more like"
"ly to climb up the ladder more quickly than their colleagues who content themselves "
"with the skills they already have.Besides, in our daily life, people who keep lear"
"ning new skills are more positive, and everyone around them must be fond of makin"
"g friends with them.";
    char *p = longsentence(string);
    printf("Longest sentence is\n%s.",p);
    return 0;
}

char *longsentence(char * string){
    char *p = strtok(string,".");
    int maxlength = strlen(p);
    char *sentence = p;
    while(1){
        p = strtok(NULL,".");
        if(p==NULL)
            break;
        if(strlen(p)>maxlength){
            maxlength = strlen(p);
            sentence = p;
        }
    }
    return sentence;
}

---

3.回文字符串

判断一个字符串是否是回文串。回文串是指从左往右以及从右往左阅读都完全相同的字符串。

输入：

"sdasfsads"

输出：

yes

题解：只需要验证两边内容一致即可

#include 
#include 
int test(char const *str);

int main(void){
    char string[] ="sdasfsads";
    int a = test(string);
    if(a==1){printf("yes");}else{printf("no");}
    return 0;
}

int test(char const *str){
    int length = strlen(str);
    if(length < 2)
        return 1;
    int p1 = 0;
    int p2 = length-1;
    length/=2;
    for(int i = 0;i<length;i++){
        if(str[p1]==str[p2]){
            p1+=1;
            p2-=1;
        }else{return 0;}
    }
    return 1;
}

---

4.粘贴一个回文字符串 ①

通过将其中两个字符串进行粘贴在一起，可以获得一个新字符串。
判断一下，能否通过这种方式粘贴两个字符串，创造出一个回文字符串呢？

题解：直接连接，用strcat()函数，再判断一下是不是回文串。

int test(char const *str){
    int length = strlen(str);
    if(length < 2)
        return 1;
    int p1 = 0;
    int p2 = length-1;
    length/=2;
    for(int i = 0;i<length;i++){
        if(str[p1]==str[p2]){
            p1+=1;
            p2-=1;
        }else{return 0;}
    }
    return 1;
}

int paste(char const *s1,char const *s2){
    char *str = malloc(sizeof(char)*20);
    strcpy(str,s1);
    int a = test(strcat(str,s2));
	free(str);
    return a;
}

---

5.粘贴一个回文字符串 ②

这一次和上一道题一样，不过我们一下子有了许许多多的字符串，现在需要考虑的是：
能否在这一大堆字符串当中挑出来其中两个，粘贴成一个回文串。
你需要探究是否可行的同时，还要将粘贴过后的字符串打印出来。

输入：

"abc";
"dcba";
"efg";
"ert";
"rty";

输出：

true
abcdcba

题解：有了前面两道题的铺垫，这样就比较容易解决这种问题了。最简单的想法当然是用两层for循环来尝试所有这些字符串，将其正反拼接一次，看看能否得到一个回文串。
再次用到前两问的函数即可。下面是完整代码:

#include 
#include 
#include 

int test(char const *str);
int stick(char const *,char const *);
char *find(char **p,int psize);

int main(void){
    char *p[5];
    p[0] = "abc";
    p[1] = "dcba";
    p[2] = "efg";
    p[3] = "ert";
    p[4] = "rty";
    char *s = find(p,5);
    if(s == NULL){printf("false");}
	else{printf("true\n%s",s);}
    return 0;
}

char *find(char **p,int psize){
    if(psize == 0){return NULL;}
    if(psize == 1){return (test(p[0]) == 1)?p[0]:NULL;}
    char *store = malloc(sizeof(char)*10);
    for(int i = 0;i < psize - 1;i++){
        for(int j = i+1;j < psize;j++){
            if(stick(p[i],p[j]) == 1){
				strcpy(store,p[i]);
				return strcat(store,p[j]);
			}
            if(stick(p[j],p[i]) == 1){
				strcpy(store,p[j]);
				return strcat(store,p[i]);
			}
        }
    }
    return NULL;
}

int test(char const *str){
    int length = strlen(str);
    if(length < 2)
        return 1;
    int p1 = 0;
    int p2 = length-1;
    length/=2;
    for(int i = 0;i<length;i++){
        if(str[p1]==str[p2]){
            p1+=1;
            p2-=1;
        }else{return 0;}
    }
    return 1;
}

int stick(char const *s1,char const *s2){
    char *str = malloc(sizeof(char)*20);
    strcpy(str,s1);

    int a = test(strcat(str,s2));

    free(str);
    return a;
}

---

6.元音字母转换

给你一个字符串，现要求你对其进行处理，使得处理后的字符串满足如下要求：
字符串里面的元音字母全部为大写；
其余字母全部为小写。

输入：

4
XYz
application
qwcvb
aeioOa

输出：

true
abcdcba

题解：只需要运用字符串查找函数strpbrk在字符串中查找即可。可以先将所有字母变为小写，再将其中的aeiou字母变为大写。变换大写小写需要用到toupper和tolower函数。

#include 
#include 
#include 

int main(void){
    int num=0;
    scanf("%d",&num);
    char* input[num];
    for(int i=0;i<num;i++){
        input[i]=malloc(sizeof(char)*50);
        scanf("%s",input[i]);
    }
    char *p=input[0];
    int ab=0;
    char const *group="aeiou";
    char const *group_upper="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    for(int i=0;i<num;i++){
        p=input[i];
        while(strpbrk(input[i],group_upper)!=NULL){
		*strpbrk(input[i],group_upper)=tolower(*strpbrk(input[i],group_upper));
        }
        while(strpbrk(input[i],group)!=NULL){

            *strpbrk(input[i],group)=toupper(*strpbrk(input[i],group));
        }
    }

    for(int i=0;i<num;i++){
        printf("%s\n",input[i]);
    }
    return 0;
}

---

本系列博客为我本人原创的学习笔记，尽量勤更新，如有错误欢迎各位大佬指出，Thanks♪(･ω･)ﾉ