c - 编写一个方法来替换 C 样式字符串中的所有空格

Question

我遇到过这个面试问题，希望得到一些帮助来理解它的解决方案:

编写一个方法，将字符串中的所有空格替换为‘%20’。

解决方案(来自论坛):

char str[]="helo b";
int length = strlen(str);
int spaceCount = 0, newLength, i = 0;

for (i = 0; i < length; i++) {
if (str[i] == ' ') {
    spaceCount++; //count spaces...
}
}

newLength = length + spaceCount * 2;   //need space for 2 more characters..
str[newLength] = '\0';

for (i = length - 1; i >= 0; i--) {
    if(str[i] == ' ') {
        str[newLength - 1] = '0'; //???
        str[newLength - 2] = '2';
        str[newLength - 3] = '%';
        newLength = newLength - 3;
    } else {
        str[newLength - 1] = str[i];
        newLength = newLength - 1;
    }
}

这个程序不适合我......我想在深入研究代码之前先了解算法。

Answer 1

那个样本坏了。

缓冲区溢出，对字符串 (strlen) 进行一次毫无意义的扫描，难以阅读。

int main() {
  char src[] = "helo b";
  int len = 0, spaces = 0;
  /* Scan through src counting spaces and length at the same time */
  while (src[len]) {
    if (src[len] == ' ')
      ++spaces;
    ++len;
  }
  /* Figure out how much space the new string needs (including 0-term) and allocate it */
  int newLen = len + spaces*2 + 1;
  char * dst = malloc(newLen);
  /* Scan through src and either copy chars or insert %20 in dst */
  int srcIndex=0,dstIndex=0;
  while (src[srcIndex]) {
    if (src[srcIndex] == ' ') {
      dst[dstIndex++]='%';
      dst[dstIndex++]='2';
      dst[dstIndex++]='0';
      ++srcIndex;
    } else {
      dst[dstIndex++] = src[srcIndex++];
    }
  }
  dst[dstIndex] = '\0';
  /* Print the result */
  printf("New string: '%s'\n", dst);
  /* And clean up */
  free(dst);
  return 0;
}