Question

我必须为一个在字符串中找到子字符串的赋值编写一些代码。

这是我的代码，我添加了评论：

// the target is the substring that we want to find in the source string
// m is the length of the target, and n is the length of the source
int contains(char target[], int m, char source[], int n) {
int flag = 0; // the source originally does not contain the target
int i;

    // go through each character of the source string
for(i = 0; i < n; i++) {
    int targetIndex = 0;
    int j;

            // check if the preceding characters of the source string are a substring
            // that matches the target string
    for(j = i; j < n && targetIndex < m; j++) {
        if(target[targetIndex] == source[j]) {
            flag = 1;
            targetIndex += 1;
        }
        else {
            flag = 0; // a letter does not match
            break;
        }
    }
}

return flag;

}

因此，当我测试此方法时，我总是返回0，我无法理解为什么如果我尝试int i = contains("potatoes", 8, "toes", 4);，则会0 我已经尝试了一些打印语句来查看它匹配的字符，它似乎只找到第一个字母"t"。

Answer 1

当你有匹配时，你需要打破外部for。

您的代码的工作方式，您可能会找到匹配项，然后再次运行外部循环并“忘记”它。

Answer 2

试试这样：

for(i = 0; i < n; i++) {
    int targetIndex = 0;
    int j;

            // check if the preceding characters of the source string are a substring
            // that matches the target string
    for(j = i; j < n && targetIndex < m; j++) {
        if(target[targetIndex] == source[j]) {
            flag = 1;
            targetIndex += 1;
        }
        else {
            flag = 0; // a letter does not match
            break;
        }
    }
  if(flag == 1)  
  {
   break;
  }
}

您可以尝试使用C的 strstr 功能，这样可以让您更轻松。

示例：

char *x= "Find the substring in this string";
char *y= "substring";
if(strstr(x, y) != NULL) {
    return true;
}

Answer 3

使用解释性注释对代码进行一些修改。

// the target is the substring that we want to find in the source string
// m is the length of the target, and n is the length of the source
int contains(char target[], int m, char source[], int n) {
int flag = 0; // the source originally does not contain the target
int i;

    // go through each character of the source string
for(i = 0; i < n; i++) {
    int targetIndex = 0;
    int j;

            // check if the preceding characters of the source string are a substring
            // that matches the target string
    for(j = i; j < n && targetIndex < m; j++) {
        if(target[targetIndex] == source[j]) {
            targetIndex += 1;
            if(targetIndex == m) { // the 'target' has been fully found
                flag = 1;
                break;
            }
        }
        else
        {
            break;
        }
    }
    if(flag == 1)  // 'target' is already found, no need to search further
    {
      break;
    }
}

return flag;
}

完全找到子字符串后，断开内部和外部循环。

EDITED：另外，根据您的功能说明，而不是int i = contains("potatoes", 8, "toes", 4);，它应该是int i = contains("toes", 4, "potatoes", 8);。

查找子字符串

3 个答案: