串：串的基本定义及简单应用魔王语言

串

串，于我的理解就是字符串
一般认为有三种存储方式：定长顺序串，堆串，块链串（个人认为比较鸡肋）。定长顺序串类似于普通字符串，同数组的大小一样最长长度固定。堆串存储在堆中，空间大小可重新分配。块链串类似于定长顺序串的链表。
定长顺序串与堆串应用较多,定长顺序串的一些操作存在截断问题

ADT String{
　　数据对象：串与字符串长度及当前串的最大长度
　　结构关系：串中的字符间存在线性关系
　　基本运算
　　initString(S)　　//初始化一个空串
　　StringCat(S,T)　　//将串T连接至串S末尾
　　StringIndexOf(S,T)　　//返回串S中第一次出现串T的位置
　　StringLastIndexOf(S,T)　　//返回串S中最后一次出现串T的位置
　　StringInsert(S,pos,T)　　//在串S的pos位置插入串T
　　StringReplace(S,from,to)　　//将串S中的第一个from子串替换为to
　　StringReplaceAll(S,from,to)　　//将串S中的所有子串from替换为to
　　StringReserve(S)　　//将串S翻转
　　SubString(S,begin,end,T)　　//从串S中begin至end取子串T
}ADT String

串的一些基本概念

空串　　　　S =""，串中没有字符，strlen(S) = 0
空格串　　　"　　"，串中的字符全为空格
子串　　　　串的一部分,设串T，串S，StringIndexOf(S,T)结果不为-1，则T为主串S的子串
前缀　　　　对于"abcde"，“a”，“ab”，“abc”，“abcd”，“abcde”都是前缀
后缀　　　　对于"abcde"，“e”，“de”，“cde”，“bcde”，“abcde”都是后缀
前缀子串　　从主串左侧第一个字符开始的所有子串
后缀子串　　在主串右侧最后一个字符结束的所有子串
串相等　　　串长相同，串中每一个位置的字符相同
模式匹配　　在主串中寻找子串首次出现位置的运算
真前/后缀（子串）　　与主串不相等

堆串的一些基本运算的实现

typedef struct{char *str;int maxlen;
}String;void initString(String *s) {s->str = (char *)malloc(sizeof(char)*STR);s->maxlen = 100;
}int StringIndexOf(String *s, char sub[]) {//返回子串开始的下标，无则返回-1char *str = s->str;int i, j;for (i = 0; i < strlen(str); i++) {j = 0;for (j = 0; j < strlen(sub); j++) {if (str[i + j] != sub[j])break;}if (j == strlen(sub)) {return i;}}return -1;
}int StringIndexOf(String *s, char c) {char *str = s->str;for (int i = 0; i < strlen(str); i++) {if (str[i] == c)return i;}return -1;
}int StringLastIndexOf(String *s, char c) {char *str = s->str;for (int i = strlen(str) - 1; i >=0; i--) {if (str[i] == c)return i;}return -1;
}void StringCat(String *s , char sub[]) {char *str = s->str;char *re = NULL;int len = strlen(str) + strlen(sub);if ( len <= s->maxlen) {strcat(str,sub);}else {re = (char *)malloc(sizeof(char)*len + 25);strcpy(re, str);strcat(re, sub);free(str);s->str = re;s->maxlen = len + 24;}
}void StringReplace(String *s, char sub[], char obj[]) {//int pos = StringIndexOf(s, sub);if (pos == -1)return;char *str = s->str;int subLen = strlen(sub), objLen = strlen(obj);char tem[125], *re = NULL;strcpy(tem, &str[pos + subLen]);int len = strlen(str) - subLen + objLen;if (len <= s->maxlen) {strcpy(str + pos, obj);strcpy(&str[strlen(str)], tem);}else {re = (char *)malloc(sizeof(char)*len + 25);strncpy(re, str, pos);strcpy(&re[pos],obj);strcpy(&re[pos+objLen], tem);free(str);s->str = re;s->maxlen = len + 24;}
}void StringReplace(String *s, int pos, int subLen, char obj[]) {//char *str = s->str;int objLen = strlen(obj);char tem[125], *re = NULL;strcpy(tem, &str[pos + subLen]);int len = strlen(str) - subLen + objLen;if (len <= s->maxlen) {strcpy(str + pos, obj);strcpy(&str[strlen(str)], tem);}else {re = (char *)malloc(sizeof(char)*len + 25);strncpy(re, str, pos);strcpy(&re[pos], obj);strcpy(&re[pos + objLen], tem);free(str);s->str = re;s->maxlen = len + 24;}
}void StringReplaceAll(String *s, char sub[], char obj[]) {int pos;int subLen = strlen(sub);while ((pos = StringIndexOf(s, sub)) != -1) {StringReplace(s, pos, subLen, obj);}
}void StringInsert(String *s , int pos, char in[]) {//从指定位置开始插入字符串char *str = s->str;char *re = NULL;char tem[125];strcpy(tem, &str[pos]);int requirement = strlen(s->str) + strlen(in);if ( requirement <= s->maxlen) {//空间足够strcpy(&str[pos], in);strcpy(&str[strlen(str)], tem);}else {re = (char *)malloc(sizeof(char)*(requirement + 25));strncpy(re, str, pos);strcpy(&(re[pos]), in);strcpy(&re[strlen(re)], tem);free(str);s->str = re;s->maxlen = requirement + 24;}
}char *SubString(String *s, int begin, int end, char *des) {strncpy(des, &(s->str)[begin], end - begin + 1);des[end - begin + 1] = 0;return des;
}char *StringReverse(char *str) {int i, len = strlen(str);char c;for (i = 0; i < len / 2; i++) {c = str[i];str[i] = str[len - i - 1];str[len - i - 1] = c;}return str;
}

魔王语言解释器基于串的实现

两条规则
(1）α->β1β2…βn
(2）(θδ1δ2…δn)->θδnθδn-1…θδ1θ

总觉得跟语法解释器有一点点相似
但是想不到方法去自定义与上述两条规则这种层次的规则

只实现了基于规则(1)的自定义规则，任意字符串可替换为任意字符串
带括号的按规则(2)去解释可嵌套括号

解释括号的算法

void EliminateBracket(String *s, int left, int right) {//left、right 左右括号的位置if (right <= left)return;char *str = s->str;char sub[NUM];SubString(s, left+2, right-1, sub);char re = str[left + 1];StringReverse(sub);char resub[NUM*2+1];int i, j;for (i = 0, j = 0; i < strlen(sub); i++) {resub[j++] = re;resub[j++] = sub[i];}resub[j++] = re;resub[j] = 0;StringReplace(s, left, right-left+1, resub);//int pos, int subLen  这个方法的第三个参数为被替换串的串长left = StringLastIndexOf(s, '('), right = StringIndexOf(s, ')');EliminateBracket(s, left, right);//递归消除括号
}

总体实现

//全局变量
int amount = 0;//规则条数
char source[NUM][LEN], object[NUM][LEN];//字符串数组，相同下标的两个source串和object串分别表示一条规则中的被替换串与用来替换的串void check_rule() {//打印规则system("cls");printf("\t\t高等规则\n");printf("\tα   →   β1β2…βm\n");printf("\t(θδ1δ2…δn)   →   θδnθδn－1… θδ1θ\n\n");printf("\t\t当前规则\n");int i = 0;while (i < amount) {printf("\t%d、\t%s   →   %s\n", i + 1, source[i], object[i]);i++;}
}int import_rule() {//从文件中读入规则int i = 0, j = 0;FILE *fp = fopen("TheLanguageOftheKingOftheDevil.rule", "r");if (fp == NULL) { printf("文件丢失！\n"); return 0; }while (fscanf(fp,"%s   →   %s", source[i++], object[j++]) != EOF);if (i != j) {printf("文件损坏！");}fclose(fp);return i-1;
}void write_rule() {//将规则写入文件int i;FILE *fp = fopen("TheLanguageOftheKingOftheDevil.rule", "w");for (i = 0; i < amount; i++) {fprintf(fp, "%s   →   %s\n", source[i], object[i]);}fclose(fp);
}void use_rule(String *s) {//使用规则int i;for (i = 0; i < amount;i++) {StringReplaceAll(s, source[i], object[i]);}EliminateBracket(s, StringLastIndexOf(s, '('), StringIndexOf(s, ')'));
}void add_rule() {char from[NUM], to[NUM];printf("请输入要转换的魔王词汇:");scanf("%s", from);printf("请输入该词汇对应的内容:");scanf("%s", to);if (confirm()) {strcpy(source[amount], from);strcpy(object[amount++], to);//保存到文件write_rule();}else {//取消printf("本次修改已取消");}
}void sort_rule() {printf("现在一共有%d条规则,如下所示:\n", amount);int i = 0, j;while (i < amount) {printf("\t%d、\t%s   →   %s\n", i + 1, source[i], object[i]);i++;}printf("输入%d个数表示当前规则的新序号，新序号-1则表示删除该条规则\n,保留的规则序号按先后顺序从小到大\n请输入:",amount);int order[NUM], cnt = 0;char Tsource[NUM][LEN], Tobject[NUM][LEN];for (i = 0; i < amount; i++) {scanf("%d", &order[i]);if (order[i] != -1)cnt++;strcpy(Tsource[i], source[i]);strcpy(Tobject[i], object[i]);}for (i = 1; i < cnt; i++) {//根据order排序for (j = 0; j < amount; j++) {if (order[j] == i)break;}strcpy(source[i - 1], Tsource[j]);strcpy(object[i - 1], Tobject[j]);}amount = cnt;write_rule();printf("修改已完成\n");
}void go_on() {printf("按任意键继续");getch();//getch();VS下可能需要两个getch()
}int confirm() {char choice[25] , flag = 1;printf("是否确认本次操作?(0/1):");do {if (flag == 0) {printf("请输入0或1表示选择\n");}scanf("%s", choice);} while (choice[0] != '0' && choice[0] != '1');return choice[0] - '0';
}int main() {//return 0;char choice[20];amount = import_rule();String ss;String *s = &ss;initString(s);while (1) {menu();scanf("%s", choice);switch (choice[0]){case '1':check_rule(); break;case '2':add_rule(); break;case '3':sort_rule(); break;case '4':printf("请输入待解释字符串:"); scanf("%s", s->str); use_rule(s); printf("结果为:%s\n",s->str); break;case '0':exit(0); break;default:printf("输入有误，请重新选择\n");break;}go_on();}return 0;
}

2018/10/13