Question

我必须使用C语言查找字符串中子字符串的计数。我正在使用函数strstr，但它只找到第一次出现。

我对算法的想法就像在字符串中搜索而strstr不返回null和在每个循环上对主字符串进行子串。我的问题是如何做到这一点？

Answer 1

您可以执行类似

的操作

int count = 0;
const char *tmp = myString;
while(tmp = strstr(tmp, string2find))
{
   count++;
   tmp++;
}

也就是说，当你得到一个结果时，再次开始在字符串的下一个位置搜索。

strstr（）不仅从字符串的开头起作用，而且从任何位置起作用。

Answer 2

是否应该使用已处理过的字符串部分？

例如，在oo， 2或3 中搜索foooo的情况下，期望的答案是什么？

如果是后者（我们允许子串重叠，答案是3），那么Joachim Isaksson suggested是正确的代码。

如果我们搜索不同的子字符串（答案应为2），请参阅下面的代码（以及在线示例here）：

char *str = "This is a simple string";
char *what = "is";

int what_len = strlen(what);
int count = 0;

char *where = str;

if (what_len) 
    while ((where = strstr(where, what))) {
        where += what_len;
        count++;
    }

Answer 3

USE KMP你可以在O（n）

中完成

int fail[LEN+1];
char s[LEN];
void getfail()
{
    //f[i+1]= max({j|s[i-j+1,i]=s[0,j-1],j!=i+1})
    //the correctness can be proved by induction
    for(int i=0,j=fail[0]=-1;s[i];i++)
    {
        while(j>=0&&s[j]!=s[i]) j=fail[j];
        fail[i+1]=++j;
        if (s[i+1]==s[fail[i+1]]) fail[i+1]=fail[fail[i+1]];//optimizing fail[]
    }
}

int kmp(char *t)// String s is pattern and String t is text!
{
    int cnt=0;
    for(int i=0,j=0;t.s[i];i++)
    {
        while(j>=0&&t.s[i]!=s[j]) j=fail[j];
        if (!s[++j])
        {
            j=fail[j];
            cnt++;
        }
    }
    return cnt;// how many times s appeared in t.
}

Answer 4

根据您是否允许重叠，结果可能会有所不同：

// gcc -std=c99
#include <stdbool.h>
#include <stdio.h>
#include <string.h>

static int
count_substr(const char *str, const char* substr, bool overlap) {
  if (strlen(substr) == 0) return -1; // forbid empty substr

  int count = 0;
  int increment = overlap ? 1 : strlen(substr);
  for (char* s = (char*)str; (s = strstr(s, substr)); s += increment)
    ++count;
  return count;
}

int main() {
  char *substrs[] = {"a", "aa", "aaa", "b", "", NULL };
  for (char** s = substrs; *s != NULL; ++s)
    printf("'%s' ->  %d, no overlap: %d\n", *s, count_substr("aaaaa", *s, true),
       count_substr("aaaaa", *s, false));
}

Output

'a' ->  5, no overlap: 5
'aa' ->  4, no overlap: 2
'aaa' ->  3, no overlap: 1
'b' ->  0, no overlap: 0
'' ->  -1, no overlap: -1

Answer 5

假设s和substr非空且非空：

/* #times substr appears in s, no overlaps */
int nappear(const char *s, const char *substr)
{
    int n = 0;
    const char *p = s;

    size_t lenSubstr = strlen(substr);

    while (*p) {
        if (memcmp(p, substr, lenSubstr) == 0) {
            ++n;
            p += lenSubstr;
        } else 
            ++p;
    }
    return n;
}

Answer 6

/* 
 * C Program To Count the Occurence of a Substring in String 
 */
#include <stdio.h>
#include <string.h>

char str[100], sub[100];
int count = 0, count1 = 0;

void main()
{
    int i, j, l, l1, l2;

    printf("\nEnter a string : ");
    scanf("%[^\n]s", str);

    l1 = strlen(str);

    printf("\nEnter a substring : ");
    scanf(" %[^\n]s", sub);

    l2 = strlen(sub);

    for (i = 0; i < l1;)
    {
        j = 0;
        count = 0;
        while ((str[i] == sub[j]))
        {
            count++;
            i++;
            j++;
        }
        if (count == l2)
        {
            count1++;                                   
            count = 0;
        }
        else
            i++;
    }    
    printf("%s occurs %d times in %s", sub, count1, str);
}

在字符串中查找子字符串的计数

6 个答案:

Output