两个整数字符串。检查第二个是否是子串

Question

#include <stdio.h>
int main() {
  int x[100], y[100], i, j, dim1, dim2, subsir=0;
  scanf("%d", &dim1);
  for(i=0; i<dim1; i++)
  {
      scanf("%d", &x[i]);
  }
  scanf("%d", &dim2);
  for(j=0; j<dim2; j++)
  {
      scanf("%d", &y[j]);
  }
  if(dim1<dim2)
  {
      printf("no");
      return 0;
  }
  for(i=0; i<dim1; i++)
    for(j=0; j<dim2; j++)
        {
        if(x[i]==y[j])
            subsir=1;
        }


  if(subsir==1)
  {
      printf("yes");
  }
  else
  {
      printf("no");
  }

  return 0;
}

考虑两个整数字符串。检查第二个字符串是否是C中第一个字符串的子字符串。（子字符串由第一个字符串的连续元素组成）

Answer 1

基于您的方法的可能实现：

// this function will return `1` in case `arr2` is contained inside `arr1`, otherwise `0`
int is_subarr(int arr1[], size_t arr1_len, int arr2[], size_t arr2_len) {
  // as you already did, we first check that arr1's len is not smaller than arr2's
  if (arr1_len < arr2_len) {
    return 0;
  }

  // useful check: in many implementations, when `arr2` is empty, they return true
  if (arr2_len == 0) {
    return 1;
  }

  // first loop: we go through all elements of `arr1` (but we stop when we have less elements left than `arr2`'s len)
  for (size_t i = 0; i < arr1_len && i < arr2_len; i += 1) {
    // temporary flag, we just pretend that `arr2` is contained, and then we check for the opposite
    int eq = 1;

    // starting with the current position `i` inside `arr1`, we start checking if `arr2` is contained
    for (size_t j = 0; j < arr2_len; j += 1) {
      // if it's not contained, we break from this inner loop, and we keep searching through `arr1`
      if (arr1[i + j] != arr2[j]) {
          eq = 0;
          break;
      }
    }

    // in case `arr2` is effectively found, we return `1`
    if (eq == 1) {
      return 1;
    }
  }

  return 0;
}

注意：“整数字符串”在 C 中称为“整数数组”；相反，“字符串”是“字符数组”。

Answer 2

您可以通过蛮力方法实现您的目标，尝试从 0 到 dim1 - dim2 的每个位置。

这是一个带有错误检查的修改版本：

#include <stdio.h>

int main() {
    int x[100], y[100], i, j, dim1, dim2;

    if (scanf("%d", &dim1) != 1 || dim1 > 100)
        return 1;
    for (i = 0; i < dim1; i++) {
        if (scanf("%d", &x[i]) != 1)
            return 1;
    }
    if (scanf("%d", &dim2) != 1 || dim2 > 100)
        return 1;

    /* no need to read second array if larger */
    if (dim1 < dim2) {
        printf("no\n");
        return 0;
    }
    for (j = 0; j < dim2; j++) {
        if (scanf("%d", &y[j]) != 1)
            return 1;
    }
    /* trying every possible sub-array start in array `x` */
    for (i = 0; i <= dim1 - dim2; i++) {
        /* compare all entries */
        for (j = 0; j < dim2; j++) {
            if (x[i + j] != y[j])
                break;
        }
        /* if above loop completed, we have a match */
        if (j == dim2) {
           printf("yes\n");
           return 0;
        }
    }
    printf("no\n");
    return 0;
}

两个整数字符串。检查第二个是否是子串

two strings of integers. Check if the second one is substring

c

string

substring