如何计算一个字符串在另一个字符串中出现的次数？ [复制]

Question

例如，我需要找出 test 出现在 ttest 中的次数，答案是 2，或者例如 world在 w1o1r1l1d 中，答案是一个。我已经写了一个找到所有可能性的代码，然后检查它是否是我正在搜索的字符串，但这太慢了。

Answer 1

我会尝试递归解决方案。

一个字母的字符串出现在另一个字符串中的次数是字符出现在那里的次数。

the number of time "r" appears in "program" is 2

一个n字母字符串在另一个字符串中出现的次数是：
(n-1)- 字符串在第一个字母的第一次匹配后出现的次数加上第一次匹配后 n 字母字符串出现的次数

the number of times "test" appears in "ttest" is
      the number of times "est" appears in "test"
    + the number of times "test" appears in "test"

---

#include <stdio.h>
#include <string.h>

int count(const char *needle, const char *stack) {
  int n = 0;
  const char *p;
  if (*stack == 0) return 0;
  if (*needle == 0) return 0;
  p = strchr(stack, *needle);
  if (needle[1] == 0) n += !!p;
  if (p) {
    n += count(needle + 1, p + 1);
    n += count(needle, p + 1);
  }
  return n;
}

int main(void) {
  const char *needle, *stack;

  needle = "a"; stack = "";
  printf("[%s] exists %d times in [%s]\n", needle, count(needle, stack), stack);

  needle = ""; stack = "a";
  printf("[%s] exists %d times in [%s]\n", needle, count(needle, stack), stack);

  needle = "a"; stack = "abracadabra";
  printf("[%s] exists %d times in [%s]\n", needle, count(needle, stack), stack);

  needle = "br"; stack = "abracadabra";
  printf("[%s] exists %d times in [%s]\n", needle, count(needle, stack), stack);

  needle = "test"; stack = "ttest";
  printf("[%s] exists %d times in [%s]\n", needle, count(needle, stack), stack);

  needle = "world"; stack = "w1o1r1l1d";
  printf("[%s] exists %d times in [%s]\n", needle, count(needle, stack), stack);

  return 0;
}