我有一个问题，就是在一段时间内找到两个字符串中最长的常用字。首先，我有一个想法，用＃34; isspace＆＃34;功能，但无法弄清楚如何找到一个常用词。然后＆＃34; strcmp＆＃34;我想到了，但到目前为止我只能比较两个字符串。我想到了一些方法来合并strcmp和isspace以便找到不同的单词，然后使用temp值来找到最长的单词，但我无法想出正确的代码。
#include <stdio.h>
int strcmp(char s[],char t[]);
void main()
{
char s[20],t[20];
printf("Type in a string s.\n");
gets(s);
printf("Type in a string t.\n");
gets( t );
printf("The result of comparison=%d\n",strcmp(s,t));
return 0;
}
int strcmp(char s[],char t[])
{
int i;
for(i=0;s[i]==t[i];i++)
if(s[i]=='\0')
return( 0 );
return(s[i]-t[i]);
}
请帮我这个。所有的想法（和代码）都受到欢迎和赞赏。提前谢谢！
编辑::
我一直在和这个人争斗一段时间，我认为我有解决方案，但这是一种非常严格的方法。该程序有一个bug，可能是数组＆＃34; ptrArray1＆＃34;但我无法修复它。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int returnArrayOfWords (char* str4Parsing, char* arrayParsed[])
{
// returns the length of array
char seps[] = " \t\n"; // separators
char *token = NULL;
char *next_token = NULL;
int i = 0;
// Establish string and get the first token:
token = strtok( str4Parsing, seps);
// While there are tokens in "str4Parsing"
while ((token != NULL))
{
// Get next token:
arrayParsed[i] = token;
//printf( " %s\n", arrayParsed[i] );//to be commented
token = strtok( NULL, seps);
i++;
}
return i;
}
void printArr(char *arr[], int n)
{
int i;
for ( i = 0; i < n; i++)
{
printf("Element %d is %s \n", i, arr[i]);
}
}
void findLargestWord(char *ptrArray1[], int sizeArr1, char *ptrArray2[], int sizeArr2)
{
int maxLength = 0;
char *wordMaxLength = NULL ;
int i = 0, j = 0;
char *w1 = NULL, *w2 = NULL; /*pointers*/
int currLength1 = 0, currLength2 = 0 ;
//printArr(&ptrArray1[0], sizeArr1);
//printArr(&ptrArray2[0], sizeArr2);
for (i = 0; i < sizeArr1; i++)
{
// to find the largest word in the array
w1 = (ptrArray1[i]); // value of address (ptrArray1 + i)
currLength1 = strlen(w1);
//printf("The word from the first string is: %s and its length is : %d \n", w1, currLength1); // check point
for (j = 0; j < sizeArr2; j++)
{
w2 = (ptrArray2[j]); // value of address (ptrArray2 + j)
currLength2 = strlen(w2);
//printf("The word from the second string is : %s and its length is : %d \n", w2, currLength2); // check point
if (strcoll(w1, w2) == 0 && currLength1 == currLength2)
// compares the strings
{
if (currLength2 >= maxLength)
// in the variable maxLength -> the length of the longest word
{
maxLength = currLength2;
wordMaxLength = w2;
printf("The largest word for now is : %s and its length is : %d \n", wordMaxLength, maxLength); // check point
}
}
}
}
printf("The largest word is: %s \n", wordMaxLength);
printf("Its length is: %d \n", maxLength);
}
int main ()
{
int n = 80; /*max number of words in string*/
char arrS1[80], arrS2[80];
char *ptrArray1 = NULL, *ptrArray2 = NULL;
int sizeArr1 = 0, sizeArr2 = 0;
// to allocate memory:
ptrArray1 = (char*)calloc(80, sizeof(char));
if(ptrArray1 == NULL)
{
printf("Error! Memory for Pointer 1 is not allocated.");
exit(0);
}
ptrArray2 = (char*)calloc(80, sizeof(char));
if(ptrArray2 == NULL)
{
printf("Error! Memory for Pointer 2 is not allocated.");
exit(0);
}
printf("Type your first string: ");
fgets(arrS1, 80, stdin);
sizeArr1 = returnArrayOfWords (arrS1, &ptrArray1); // sizeArr1 = number of elements in array 1
printf("Type your second string: ");
fgets(arrS2, 80, stdin);
sizeArr2 = returnArrayOfWords (arrS2, &ptrArray2); // sizeArr2 = number of elements in array 2
findLargestWord(&ptrArray1, sizeArr1, &ptrArray2, sizeArr2);
free(ptrArray1);
free(ptrArray2);
return 0;
}
我也尝试使用后两个发布的解决方案，但是如下所述，我有使用它们的问题。
对我的代码提供任何帮助，解决后一种解决方案的问题或提出新的解决方案都是值得欢迎的。提前谢谢大家！
PS。如果我的代码位置不好，我很抱歉。我使用这个展示位置仍然不太好。
答案 0 :(得分：1)
有很多方法可以解决这个问题。下面，指向其中一个字符串中每个字符的指针用于使用strchr搜索另一个字符以匹配字符。找到匹配字符后，比较循环会运行推进每个指针，以设计存在的公共子字符串的长度（如果有）。
例程，匹配字符，检查子字符串长度，重复，继续如此长，因为strchr返回一个有效指针。每次找到一个更长的子字符串时，max长度会更新以返回并且子字符串存在被复制到r strncpy和 nul-terminated ，以便最长公共字符串的文本可供调用函数main使用。
这是一种相当强力的方法，可能会有一些额外的调整来提高效率。功能本身是：
/** return length of longest common substring in 'a' and 'b'.
* by searching through each character in 'a' for each match
* in 'b' and comparing substrings present at each match. the
* size of the longest substring is returned, the test of the
* longest common substring is copied to 'r' and made available
* in the calling function. (the lengths should also be passed
* for validation, but that is left as an exercise)
*/
size_t maxspn (const char *a, const char *b, char *r)
{
if (!a||!b||!*a||!*b) return 0; /* valdate parameters */
char *ap = (char *)a; /* pointer to a */
size_t max = 0; /* max substring char */
for (; *ap; ap++) { /* for each char in a */
char *bp = (char *)b; /* find match in b with strchr */
for (; *bp && (bp = strchr (bp, *ap)); bp++) {
char *spa = ap, *spb = bp; /* search ptr initialization */
size_t len = 0; /* find substring len */
for (; *spa && *spb && *spa == *spb; spa++, spb++) len++;
if (len > max) { /* if max, copy to r */
strncpy (r, ap, (max = len));
r[max] = 0; /* nul-terminate r */
}
}
}
return max;
}
返回长度max，然后在函数执行期间更新r会导致r保存与最长子字符串匹配关联的字符串。
由于其安全风险，其他改进措施是删除在C11中删除的gets而不弃用。任何理智的编码器都不应该使用它（应该覆盖我们大约40％的人）。将剩余的位放在一起，可以使用一小段测试代码：
#include <stdio.h>
#include <string.h>
#define MAXC 128
size_t maxspn (const char *a, const char *b, char *r);
void rmlf (char *s);
int main (void) {
char res[MAXC] = "", s[MAXC] = "", t[MAXC] = "";
printf ("Type in a string 's': ");
if (!fgets (s, MAXC, stdin)) { /* validate 's' */
fprintf (stderr, "error: invalid input for 's'.\n");
return 1;
}
rmlf (s); /* remove trailing newline */
printf ("Type in a string 't': ");
if (!fgets (t, MAXC, stdin)) { /* validate 't' */
fprintf (stderr, "error: invalid input for 's'.\n");
return 1;
}
rmlf (t); /* remove trailing newline */
/* obtain longest commons substring between 's' and 't' */
printf ("\nThe longest common string is : %zu ('%s')\n",
maxspn (s, t, res), res);
return 0;
}
/** return length of longest common substring in 'a' and 'b'.
* by searching through each character in 'a' for each match
* in 'b' and comparing substrings present at each match. the
* size of the longest substring is returned, the test of the
* longest common substring is copied to 'r' and made available
* in the calling function. (the lengths should also be passed
* for validation, but that is left as an exercise)
*/
size_t maxspn (const char *a, const char *b, char *r)
{
if (!a||!b||!*a||!*b) return 0; /* valdate parameters */
char *ap = (char *)a; /* pointer to a */
size_t max = 0; /* max substring char */
for (; *ap; ap++) { /* for each char in a */
char *bp = (char *)b; /* find match in b with strchr */
for (; *bp && (bp = strchr (bp, *ap)); bp++) {
char *spa = ap, *spb = bp;
size_t len = 0; /* find substring len */
for (; *spa && *spb && *spa == *spb; spa++, spb++) len++;
if (len > max) { /* if max, copy to r */
strncpy (r, ap, (max = len));
r[max] = 0; /* nul-terminate r */
}
}
}
return max;
}
/** remove trailing newline from 's'. */
void rmlf (char *s)
{
if (!s || !*s) return;
for (; *s && *s != '\n'; s++) {}
*s = 0;
}
示例使用/输出
$ ./bin/strspn
Type in a string 's': a string with colors123456789 all blue
Type in a string 't': a string without colors1234567890 all red
The longest common string is : 16 (' colors123456789')
或另一个可能更容易想象的内容：
$ ./bin/strspn
Type in a string 's': green eel
Type in a string 't': cold blue steel
The longest common string is : 3 ('eel')
查看代码并与其他答案进行比较。如果您有任何其他问题，请与我们联系。还有一些其他验证应该添加到确保文本不会写入缓冲区的末尾等等。希望这将提供一些帮助或替代方法。
其他子字符串
为了确保你和我看到同样的事情，我在下面列出了其他使用示例。没有错误，代码按预期执行。如果您在修改代码时遇到问题，请告诉我您尝试做什么，我可以提供帮助。我的代码中的每个指针增量都经过验证。如果您更改了有关指针增量或 nul-termination 的任何内容，除非您考虑验证中的更改，否则代码将无效。
$ ./bin/strspn
Type in a string 's': 1
Type in a string 't':
The longest common string is : 0 ('')
$ ./bin/strspn
Type in a string 's': A man a plan a canal panama
Type in a string 't': a man a plan a river panama
The longest common string is : 14 (' man a plan a ')
$ ./bin/strspn
Type in a string 's': this is my favorite string
Type in a string 't': this is my favoritist string
The longest common string is : 18 ('this is my favorit')
$ ./bin/strspn
Type in a string 's': not the same until here
Type in a string 't': cant be equal till here
The longest common string is : 6 ('l here')
$ ./bin/strspn
Type in a string 's': some str with ten in the middle
Type in a string 't': a string often ignorded
The longest common string is : 5 ('ten i')
最长的常用词
好的，在我最终了解您要完成的工作后，您可以通过标记化选择两个字符串's'和't'之间的最长公用字 每个字符串都带有strtok，在单独的指针数组中保存指向每个字符串中每个字的指针，然后简单地遍历指针数组以选择最长的常用字（如果是同一个多个常用字，则为第1个字）长度）。您需要的只是以下简单的东西。
注意 strtok修改字符串's'和't'，因此如果您需要保留原件，请复制一份。
/** return length of longest common word in 'a' and 'b'.
* by tokenizing each word in 'a' & 'b' and iterating over
* each, returning the length of the logest match, and updating
* 'r' to contain the longest common word.
*/
size_t maxspnwhole (char *a, char *b, char *r)
{
if (!a||!b||!*a||!*b) return 0; /* valdate parameters */
char *arra[MAXC] = {NULL}, *arrb[MAXC] = {NULL};
char *ap = a, *bp = b; /* pointers to a & b */
char *delim = " .,-;\t\n"; /* word delimiters */
size_t i, j, len, max, na, nb; /* len, max, n-words */
len = max = na = nb = 0;
/* tokenize both strings into pointer arrays */
for (ap = strtok (a, delim); ap; ap = strtok (NULL, delim))
arra[na++] = ap;
for (bp = strtok (b, delim); bp; bp = strtok (NULL, delim))
arrb[nb++] = bp;
for (i = 0; i < na; i++) /* select longest common word */
for (j = 0; j < nb; j++)
if (*arra[i] == *arrb[j]) /* 1st chars match */
if (!strcmp (arra[i], arrb[j])) { /* check word */
len = strlen (arra[i]);
if (len > max) { /* if longest */
max = len; /* update max */
strcpy (r, arra[i]); /* copy to r */
}
}
return max;
}
将其与其他代码集成，您可以比较结果：
#include <stdio.h>
#include <string.h>
#define MAXC 128
size_t maxspn (const char *a, const char *b, char *r);
size_t maxspnwhole (char *a, char *b, char *r);
void rmlf (char *s);
int main (void) {
char res[MAXC] = "", s[MAXC] = "", t[MAXC] = "";
printf ("Type in a string 's': ");
if (!fgets (s, MAXC, stdin)) { /* validate 's' */
fprintf (stderr, "error: invalid input for 's'.\n");
return 1;
}
rmlf (s); /* remove trailing newline */
printf ("Type in a string 't': ");
if (!fgets (t, MAXC, stdin)) { /* validate 't' */
fprintf (stderr, "error: invalid input for 's'.\n");
return 1;
}
rmlf (t); /* remove trailing newline */
/* obtain longest commons substring between 's' and 't' */
printf ("\nThe longest common string is : %zu ('%s')\n",
maxspn (s, t, res), res);
/* obtain longest commons word between 's' and 't' */
printf ("\nThe longest common word is : %zu ('%s')\n",
maxspnwhole (s, t, res), res);
return 0;
}
/** return length of longest common word in 'a' and 'b'.
* by tokenizing each word in 'a' & 'b' and iterating over
* each, returning the length of the logest match, and updating
* 'r' to contain the longest common word.
*/
size_t maxspnwhole (char *a, char *b, char *r)
{
if (!a||!b||!*a||!*b) return 0; /* valdate parameters */
char *arra[MAXC] = {NULL}, *arrb[MAXC] = {NULL};
char *ap = a, *bp = b; /* pointers to a & b */
char *delim = " .,-;\t\n"; /* word delimiters */
size_t i, j, len, max, na, nb; /* len, max, n-words */
len = max = na = nb = 0;
/* tokenize both strings into pointer arrays */
for (ap = strtok (a, delim); ap; ap = strtok (NULL, delim))
arra[na++] = ap;
for (bp = strtok (b, delim); bp; bp = strtok (NULL, delim))
arrb[nb++] = bp;
for (i = 0; i < na; i++)
for (j = 0; j < nb; j++)
if (*arra[i] == *arrb[j])
if (!strcmp (arra[i], arrb[j])) {
len = strlen (arra[i]);
if (len > max) {
max = len;
strcpy (r, arra[i]);
}
}
return max;
}
/** return length of longest common substring in 'a' and 'b'.
* by searching through each character in 'a' for each match
* in 'b' and comparing substrings present at each match. the
* size of the longest substring is returned, the test of the
* longest common substring is copied to 'r' and made available
* in the calling function. (the lengths should also be passed
* for validation, but that is left as an exercise)
*/
size_t maxspn (const char *a, const char *b, char *r)
{
if (!a||!b||!*a||!*b) return 0; /* valdate parameters */
char *ap = (char *)a; /* pointer to a */
size_t max = 0; /* max substring char */
for (; *ap; ap++) { /* for each char in a */
char *bp = (char *)b; /* find match in b with strchr */
for (; *bp && (bp = strchr (bp, *ap)); bp++) {
char *spa = ap, *spb = bp;
size_t len = 0; /* find substring len */
for (; *spa && *spb && *spa == *spb; spa++, spb++) len++;
if (len > max) { /* if max, copy to r */
strncpy (r, ap, (max = len));
r[max] = 0; /* nul-terminate r */
}
}
}
return max;
}
/** remove trailing newline from 's'. */
void rmlf (char *s)
{
if (!s || !*s) return;
for (; *s && *s != '\n'; s++) {}
*s = 0;
}
示例使用/输出
$ ./bin/strlongestcmn
Type in a string 's': I have a huge boat.
Type in a string 't': I have a small boat.
The longest common string is : 9 ('I have a ')
The longest common word is : 4 ('have')
仔细看看，如果您有任何其他问题，请告诉我。
答案 1 :(得分：-1)
此示例打印给定两个输入字符串的最长子字符串。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int *lcommon(char *str1, char *str2) {
int strlen1 = (unsigned) strlen(str1);
int strlen2 = (unsigned) strlen(str2);
int i, j, k;
int longest = 0;
int **ptr = malloc(2 * sizeof(int *));
static int *ret;
ret = calloc((unsigned) strlen1 + 1, sizeof(int));
for (i = 0; i < 2; i++)
ptr[i] = calloc((unsigned) strlen2, sizeof(int));
k = 0;
for (i = 0; i < strlen1; i++) {
memcpy(ptr[0], ptr[1], strlen2 * sizeof(int));
for (j = 0; j < strlen2; j++) {
if (str1[i] == str2[j]) {
if (i == 0 || j == 0) {
ptr[1][j] = 1;
} else {
ptr[1][j] = ptr[0][j-1] + 1;
}
if (ptr[1][j] > longest) {
longest = ptr[1][j];
k = 0;
ret[k++] = longest;
}
if (ptr[1][j] == longest) {
ret[k++] = i;
ret[k] = -1;
}
} else {
ptr[1][j] = 0;
}
}
}
for (i = 0; i < 2; i++)
free(ptr[i]);
free(ptr);
ret[0] = longest;
return ret;
}
int main(int argc, char *argv[]) {
int i, longest, *ret;
if (argc != 3) {
printf("usage: longest-common-substring string1 string2\n");
exit(1);
}
ret = lcommon(argv[1], argv[2]);
if ((longest = ret[0]) == 0) {
printf("There is no common substring\n");
exit(2);
}
i = 0;
while (ret[++i] != -1) {
printf("%.*s\n", longest, &argv[1][ret[i]-longest+1]);
}
exit(0);
}
<强>测试
$ ./a.out computerprogramming javaprogrammer
programm
您可以阅读有关问题here的更多信息。
您还可以使用交互式程序在控制台中编写字符串：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int *lcommon(char *str1, char *str2) {
int strlen1 = (unsigned) strlen(str1);
int strlen2 = (unsigned) strlen(str2);
int i, j, k;
int longest = 0;
int **ptr = malloc(2 * sizeof(int *));
static int *ret;
ret = calloc((unsigned) strlen1 + 1, sizeof(int));
for (i = 0; i < 2; i++)
ptr[i] = calloc((unsigned) strlen2, sizeof(int));
k = 0;
for (i = 0; i < strlen1; i++) {
memcpy(ptr[0], ptr[1], strlen2 * sizeof(int));
for (j = 0; j < strlen2; j++) {
if (str1[i] == str2[j]) {
if (i == 0 || j == 0) {
ptr[1][j] = 1;
} else {
ptr[1][j] = ptr[0][j - 1] + 1;
}
if (ptr[1][j] > longest) {
longest = ptr[1][j];
k = 0;
ret[k++] = longest;
}
if (ptr[1][j] == longest) {
ret[k++] = i;
ret[k] = -1;
}
} else {
ptr[1][j] = 0;
}
}
}
for (i = 0; i < 2; i++)
free(ptr[i]);
free(ptr);
ret[0] = longest;
return ret;
}
int main(int argc, char *argv[]) {
int i, longest, *ret;
if (argc != 3) {
//printf("usage: longest-common-substring string1 string2\n");
char s[20], t[20];
printf("Type in a string s.\n");
fgets(s, 20, stdin);
printf("Type in a string t.\n");
fgets(t, 20, stdin);
ret = lcommon(s, t);
if ((longest = ret[0]) == 0) {
printf("There is no common substring\n");
exit(2);
}
i = 0;
while (ret[++i] != -1) {
printf("%.*s\n", longest, &s[ret[i] - longest + 1]);
}
//printf("The result of comparison=%d\n", strcmp(s, t));
exit(0);
} else { }
ret = lcommon(argv[1], argv[2]);
if ((longest = ret[0]) == 0) {
printf("There is no common substring\n");
exit(2);
}
i = 0;
while (ret[++i] != -1) {
printf("%.*s\n", longest, &argv[1][ret[i] - longest + 1]);
}
exit(0);
}
<强>测试
Type in a string s.
string1
Type in a string t.
string2
string