替换字符串里的空格

请实现一个函数，将一个字符串中的空格替换成“%20”。例如，当字符串为We Are Happy.则经过替换之后的字符串为We%20Are%20Happy。

void replaceSpace(char *str, int length) {
    if (str == NULL || length < 1)
    {
        return;
    }
    int spaceCount = 0;
    for (int i = 0; i < length; i++)
    {
        if (str[i] == ' ')
        {
            ++spaceCount;
        }
    }

    int destLen = length + 2 * spaceCount;
    int j = destLen - 1;
    for (int i = length - 1; i >= 0; i--)
    {
        if (str[i] == ' ')
        {
            str[j] = '0';
            str[--j] = '2';
            str[--j] = '%';
        }
        else
        {
            str[j] = str[i];
        }
        --j;
    }
}

测试

void main() {

    char str[] = "Hello world";
    
    //char* str = "Hello world";  //不能这样用，char*指向的字符串不可修改

    replaceSpace(str, strlen(str) + 1);

    cout << str << endl;    

    getchar();
}

相关
1、C语言中"\0"和空格的区别
http://leoenglish.blog.163.com/blog/static/17503198520112945739709/
2，char *a 和 char a[]的区别
https://www.zhihu.com/question/26117197