最近我在面试中被要求将字符串“aabbbccccdddddd”转换为“a2b3c4d5”。目标是用单次出现和重复计数替换每个重复的字符。这里 'a' 在输入中重复了两次，因此我们必须在输出中将其写为 'a2'。我还需要编写一个函数来将格式反转回原始格式(例如 从字符串“a2b3c4d5”到“aabbbccccddddd”)。我可以自由使用 C 或 C++。我写了下面的代码，但是面试官似乎对此不太满意。他让我尝试一种比这更聪明的方法。

在下面的代码中，我使用 formatstring() 通过添加重复计数来消除重复的字符，并使用 reverseformatstring() 转换回原始字符串。

void formatstring(char* target, const char* source) {
  int charRepeatCount = 1;
  bool isFirstChar = true;
  while (*source != '\0') {
    if (isFirstChar) {
      // Always add the first character to the target
      isFirstChar = false;
      *target = *source;
      source++; target++;
    } else {
      // Compare the current char with previous one,
      // increment repeat count
      if (*source == *(source-1)) {
        charRepeatCount++;
        source++;
      } else {
        if (charRepeatCount > 1) {
          // Convert repeat count to string, append to the target
          char repeatStr[10];
          _snprintf(repeatStr, 10, "%i", charRepeatCount);
          int repeatCount = strlen(repeatStr);
          for (int i = 0; i < repeatCount; i++) {
            *target = repeatStr[i];
            target++;
          }
          charRepeatCount = 1; // Reset repeat count
        }
        *target = *source;
        source++; target++;
      }
    }
  }
  if (charRepeatCount > 1) {
    // Convert repeat count to string, append it to the target
    char repeatStr[10];
    _snprintf(repeatStr, 10, "%i", charRepeatCount);
    int repeatCount = strlen(repeatStr);
    for (int i = 0; i < repeatCount; i++) {
      *target = repeatStr[i];
      target++;
    }
  }
  *target = '\0';
}

void reverseformatstring(char* target, const char* source) {
  int charRepeatCount = 0;
  bool isFirstChar = true;
  while (*source != '\0') {
    if (isFirstChar) {
      // Always add the first character to the target
      isFirstChar = false;
      *target = *source;
      source++; target++;
    } else {
      // If current char is alpha, add it to the target
      if (isalpha(*source)) {
        *target = *source;
        target++; source++;
      } else {
        // Get repeat count of previous character
        while (isdigit(*source)) {
          int currentDigit = (*source) - '0';
          charRepeatCount = (charRepeatCount == 0) ?
              currentDigit : (charRepeatCount * 10 + currentDigit);
          source++;
        }
        // Decrement repeat count as we have already written
        // the first unique char to the target
        charRepeatCount--; 
        // Repeat the last char for this count
        while (charRepeatCount > 0) {
          *target = *(target - 1);
          target++;
          charRepeatCount--;
        }
      }
    }
  }
  *target = '\0';
}

我没有发现上述代码有任何问题。还有其他更好的方法吗？

最佳答案

方法/算法很好，也许您可​​以稍微改进和缩小代码(通过做一些更简单的事情，没有必要以过于复杂的方式解决这个问题)。并选择一种真正有意义的缩进样式。

A C 溶液:

void print_transform(const char *input)
{
    for (const char *s = input; *s;) {
        char current = *s;
        size_t count = 1;
        while (*++s == current) {
            count++;
        }

        if (count > 1) {
            printf("%c%zu", current, count);
        } else {
            putc(current, stdout);
        }
    }

    putc('\n', stdout);
}

(这可以很容易地修改，以便它返回转换后的字符串，或者将其写入足够长的缓冲区。)

C++ 解决方案:

std::string transform(const std::string &input)
{
    std::stringstream ss;
    std::string::const_iterator it = input.begin();

    while (it != input.end()) {
        char current = *it;
        std::size_t count = 1;
        while (++it != input.end() && *it == current) {
            count++;
        }

        if (count > 1) {
            ss << current << count;
        } else {
            ss << current;
        }
    }

    return ss.str();
}

关于c++ - 使用 C/C++ 进行字符串格式化，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/19607062/