2.1.求next和nextval的实现
代码：
int next_one(char *str, int len) {int result = 1;if(len == 1 || len == 0) return len;for (size_t i = 1; i < len; i++){ if(compare(str, str+len-i, i)) {result = i+1;//break;}}return result;
}int next(char *str, int *arr) {int count = size(str);if(count == 1) {arr[0] = 0; return count;}for (size_t i = 0; i < count; i++)//i = len{arr[i] = next_one(str, i);}return count;
}int nextval(char *str, int *arr) {int count = size(str);int k = 0;if(count == 1) {arr[0] = 0; return count;}for (size_t i = 0; i < count; i++)//i = len{k = next_one(str, i);//if(str[i] == str[k-1]) {arr[i] = 0; } else {arr[i] = k;}}return count;
}int compare(char *src_str, char *dist_str, int len) {for(int i = 0; i < len; i++) {if(src_str[i] != dist_str[i]) return 0;}return 1;
}
输出结果：
2.2展现t和p的匹配过程
- p的nextval值
- KMP的匹配过程
代码：
int KMP(char *src, char *pat) {int len_s = size(src);int len_p = size(pat);int next[100];nextval(pat, next);int j = 0;for(int i = 0; i < len_s; i++) {if(src[i] == pat[j]) {j++;} else {j = next[j];}printf("i = %d, j = %d\n", i, j);if(j == len_p) return i-len_p+1;}return -1;
}
输出结果：
2.4 广义表的定义
代码：
#include "stdio.h"typedef char* AtomType;
typedef enum{ATOM, LIST} ElemTag;
typedef struct
{ElemTag tag;union node{AtomType atom;struct{struct GLnode *head,*tail;}ptr;};
}GLnode;void init(GLnode *node);
void head(AtomType str);
void tail();
void traverse(GLnode *node);void main() {GLnode node;init(&node);//traverse(node);}void init(GLnode *node) {GLnode glnode = *node;glnode.tag=1;GLnode head;head.tag = 0;head.atom = "apple\0";GLnode tail;//tail listtail.tag = 1;glnode.ptr.head = &head;glnode.ptr.tail = &tail;GLnode head1,tail1;head1.tag = 1;tail1.tag = 0;tail1.atom = "pear\0";tail.ptr.head = &head1;tail.ptr.tail = &tail1;GLnode head2,tail2;head2.tag = 0;tail2.tag = 1;head2.atom = "orange\0";head1.ptr.head = &head2;head1.ptr.tail = &tail2;GLnode head3,tail3;head3.tag = 0;tail3.tag = 0;head3.atom = "strawberry\0";tail3.atom = "banana\0";head2.ptr.head = &head3;head2.ptr.tail = &tail3;traverse(&glnode); }void head(GLnode node, AtomType str) {if (node.tag){/* code */}}void traverse(GLnode *node) {if(!node->tag) {printf("The elem is %s\n", node->atom);} else {traverse(node->ptr.head);traverse(node->ptr.tail);}
}
3.1 统计字符串中的个数
分析：
1存放字符的数据结构需要动态地添加结点，所以选择链表保存；
2遍历字符串，然后判断链表中是否已经存在，如果不存在，新添加一个结点插入到队列；
如果存在，结点中地统计变量加1.
3 最后把链表地内容输入到文件中。
代码：
#include "stdio.h"typedef struct
{char ch;int count;struct lnode *next;
}lnode;void init(lnode *list);
void count(lnode *list, char *str);
void insert(lnode *list, char ch);
void print(lnode *list);
void write_file(lnode *list);
void num_to_str(int num, char *str);
void main() {lnode list;char *str = "ahfAJOFJIOhoahf832977689 ofj___";init(&list);count(&list, str);//print(&list);write_file(&list);}void init(lnode *list) {list->next = NULL;
}void count(lnode *list, char *str) {lnode *node = list;while (*str){if(*str >= 'A' && *str <= 'Z' || *str >= 'a' && *str <= 'z' || *str >= '0' && *str <= '9') {while (node->next){node = node->next;if(node->ch == *str) {node->count++;break;}}if(!node->next) {lnode *l_node = (lnode *) malloc (sizeof(lnode));l_node->ch = *str;l_node->count = 1;node->next = l_node;l_node->next = NULL;}}str++;node = list;}}void print(lnode *list) {while (list->next){list = list->next;printf("%c", list->ch);}}void write_file(lnode *list) {FILE *file = fopen("count.txt", "a");while (list->next){list = list->next;char str[20];str[0] = list->ch;str[1] = ':';char *p = str+2;num_to_str(list->count, p);fputs(str, file);}fclose(file);
}void num_to_str(int num, char *str) {int count = 0;for(int i = 0; num != 0; i++) {str[i] = num % 10 + '0';num = num / 10;count++;}str[count] = '\n';str[count+1] = 0;
}
输出结果：
3.2 递归实现字符串逆序
代码：
#include "stdio.h"void reverse(char *str, int i, int j);
void swap(char *a, char *b);
int size(char *str);
void main() {char *str = "abcd";int i = 0;int j = size(str) - 1;reverse(str, i, j);}void reverse(char *str, int i, int j) {if(i > j) return;swap(&str[i], &str[j]);reverse(str, i+1, j-1);
}void swap(char *a, char *b) {char temp;temp = *a;*a = *b;*b = temp;
}int size(char *str) {int length = 0;while(*str++) {length++;}return length;
}
输出结果：
3.3 像字符串中插入字符串
分析：
s0 n-1 n s_len
t0 t_len
s0 n-1 n t_len+n-1 s_len+t_len
1.n = s_len-1
2.n<s_len-1, 先搬n到s_len-1的元素到n+t_len, 直到s_len+t_len-1.
3.再填充t0-n, 从n到n+t_len-1.
代码：
#include "stdio.h"void insert(char *s, char *t, int n);
int size(char *str);
void main() {char s[200] = "abcdefg";char t[200] = "xyz";int n = 3;insert(s, t, 3);printf("The string is %s", s);}void insert(char *s, char *t, int n) {int s_len = size(s);int t_len = size(t);s[s_len+t_len] = 0;for(int i = s_len-n-1; i >= 0; i--) {s[n+t_len+i] = s[n+i];}for(int i = 0; i < t_len; i++) {s[n+i] = t[i];}
}int size(char *str) {int length = 0;while(*str++) {length++;}return length;
}
输出结果：
3.4.分S1为S1和S2.
分析：S1 len1
S2 len2 = n
S3 len3 = len1-n.
代码：
#include "stdio.h"void format(char *s1, char *s2, char *s3, int n);
int size(char *str);
void main() {char s1[200] = "abcdefgadfadfad";char s2[200];char s3[200];int n = 10;format(s1, s2, s3, n);printf("The string is %s,%s,%s", s1, s2, s3);}void format(char *s1, char *s2, char *s3, int n) {int count = size(s1);int j = 0;for(int i = 0; i < count; i++) {if(i < n) {s2[i] = s1[i];} else {s3[j++] = s1[i];}}s2[n] = 0;s3[j] = 0;
}int size(char *str) {int length = 0;while(*str++) {length++;}return length;
}
输出结果：
3.5. 求解二维数组是否有重复元素
代码：
#include "stdio.h"int same_elem_matrix(int *a[3], int m, int n);void main() {int a1[3] = {1,3,5};int a2[3] = {2,4,6};int a3[3] = {25,7,9};int *a[3] = {&a1,&a2,&a3};same_elem_matrix(a, 3, 3);}int same_elem_matrix(int *a[3], int m, int n) {int **p = a;for(int i = 0; i < m*n; i++) {for(int j = 0; j < 3; j++) {for(int k = 0; k < 3; k++) {if(i/m == j && i%n == k)break;if(*(*(p+i/m)+i%n) == a[j][k]) {printf("yes! the matrix has same elem.");return 1;}}}}printf("no! the matrix has no same elem."); return 0;
}
输出结果：
3.6. 正负数分离
分析：
申请a[n]，b[n]记录A[n]的正负分布情况。
代码：
#include "stdio.h"void check_number(int *A, int length);void main() {int a[10] = {-1,-3,5,2,-5,9,2,3,-10,-29};check_number(a, 10);for(int i = 0; i < 10; i++) {printf("%d,", a[i]);}
}void check_number(int *A, int length) {int b[10],c[10];int j = 0;int k = 0;for(int i = 0; i < length; i++) {if(A[i] >= 0) {b[j++] = A[i];} else {c[k++] = A[i];}}k = 0;for(int i = 0; i < length; i++) {if(i < j)A[i] = b[i];else A[i] = c[k++];}
}
输出结果：
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