cracking the coding interview No1.5

最新推荐文章于 2016-10-31 00:39:47 发布

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最新推荐文章于 2016-10-31 00:39:47 发布

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分类专栏： cracking the coding interview

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本文介绍了一种基本的字符串压缩方法，该方法通过计算连续字符的数量来进行压缩。如果压缩后的字符串长度不小于原始字符串，则返回原始字符串。文章提供了一个具体的C/C++实现示例。

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1.5 Implement a method to perform basic string compression using the counts of repeated characters.For example.

For example,the string aabcccccaaa would become a2b1c5a3.If the “compressed”string would not become smaller

than the original string,your method should return the original string.

void change(char *str)
{
	if (str == NULL)
		return;
	int count = 0,length = strlen(str);
	int k = 0;
	char c = str[0];
	char *laststring = new char[length+1];
	
	for (int i = 0;i < length;i++)
	{
		if (c == str[i])
		{
			count++;
		}
		else
		{
			laststring[k++] = c;
			laststring[k++] = count+'0';
			c = str[i];
			count = 0;
		}	

	}
	laststring[k] = '\0';
	if (k > length)
	{
		printf("%s\n",str);
	}
	else
	{
		printf("%s\n",laststring);
	}
}