如何压缩一串数字 [英] how to compress a string of numbers

问题描述

大家好。

我一直试图想出一种方法来压缩（压缩）只包含数字的大字符串

是否有可能将每个字符串组合起来几个数字并用8位代表它们而不是让每个角色占用整个八位字节





这是我的输入样本：



543216548798654321320000654897986321



每两个数字应该占用8位



谢谢你

解决方案

如果你真的确定它只是十进制数字你可以像这样简单：

  char  digit1 = sample [i]; 
  int  value1 = digit1  -  '  0' ;  //  介于0到9之间 
 
  char  digit0 = sample [i + 1]; 
  int  value0 = digit0  -  '  0' ;  //  介于0和9之间 
  int  number = value1 *  10  + value0; 
 
 byteBuffer [n] =（字节）数; 
 

确保数值低于128（7位），因此您可以省略1位并将值保存为7位。但这是高级的东西，你应该像来自微软这样理解它。

  unsigned   int  i; 
  const   char  * s =   123465789; 
 vector< char> NUM; 
  for （i =  0 ; i< strlen（s）; i + =  2 ）
 {
 num.push_back（（s [i]  -  '  0'）+（s [i + 1]  -  '  0' ）*  10 ）; 
} 

就像这样的工作？

将向量num的大小是字符串s的一半大小？ ？

hello everyone
I've been trying to come up with a way to condense (compress) a large string that only contains number
is it possible to combining each couple of numbers and represent them in 8 bits instead of having each character take up the entire octet


Here is my input sample:

"543216548798654321320000654897986321"

each two numbers should take up 8 bits

thank you

解决方案

if you are REALLY SURE, that it will only be decimal numbers you can go as easy as this:

char digit1 = sample[i];
int value1 = digit1 - '0'; //is between 0 and 9

char digit0 = sample[i+1];
int value0 = digit0 - '0'; //is between 0 and 9
int number = value1 * 10 + value0;

byteBuffer[n] = (byte) number;

The value number is ensured to be below 128 (7 bits), so you can spare out 1 bit and save the value in 7 bits. But that is advanced stuff and you should understand it like here from Microsoft fine explained.

unsigned int i;
	const char *s = "123465789";
	vector<char> num;
	for (i=0;i<strlen(s);i+=2)
	{
		num.push_back((s[i]-'0') + (s[i+1]-'0')*10);
	}

would something like this work ?
will the size of the vector num be half the size of the string s ??

这篇关于如何压缩一串数字的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！