【C语言】 leetcode 165. Compare Version Number

题目：

Compare two version numbers version1 and version2.

If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.

You may assume that the version strings are non-empty and contain only digits and the 

.

 character.

The 

.

 character does not represent a decimal point and is used to separate number sequences.

For instance, 

2.5

 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.

Here is an example of version numbers ordering:

0.1 < 1.1 < 1.2 < 13.37      

题目解析：

版本号以小数点作为分隔，1.2<1.05 （第一版第二次和第一版第五次，所以1.05大） 1.2=1.20  

题目解法

解法一：

char* zerospace(char*str)
{
	int i = 0;
	if (str[0]=='.'||str[0] == '\0')
	{
		return str;
	}
	else
	{
		while (str[i] == '0' || str[i] == ' ')
		{
			i++;
		}
		return str + i;
	}
}

//返回'.'或者'\0'的位置
int locatepoint(char*version)
{
	int i = 0;
	while (version[i])
	{
		if (version[i] == '.')
		{
			return i;
		}
		i++;
	}
	return i;
}

int compareVersion(char* version1, char* version2) {
	int point1;
	int point2;
	int i = 0;

	while (version1[0] || version2[0])
	{
		//去除整数部分左侧的'0'和' '
		version1 = zerospace(version1);
		version2 = zerospace(version2);

		//获取'.'或者'\0的位置
		point1 = locatepoint(version1);
		point2 = locatepoint(version2);

		//比较小数点左侧大小，可以先直接比较长度
		if (point1>point2)
			return 1;
		else if (point1<point2)
			return -1;
		else
		{
			for (int i = 0; i<point1; i++)
			{
				if (version1[i]>version2[i]) return 1;
				else if (version1[i]<version2[i]) return -1;
			}
			if (!version1[point1])
				version1 = version1 + point1;
			else
				version1 = version1 + point1 + 1;

			if (!version2[point2])
				version2 = version2 + point2;
			else
				version2 = version2 + point2 + 1;
		}
	}
	return 0;
}           

解法二：

int compareVersion(char* version1, char* version2) {
	int lenversion1 = strlen(version1);
	int lenversion2 = strlen(version2);
	int i = 0;
	int j = 0;
	int num1 = 0;
	int num2 = 0;

	while (i < lenversion1 || j < lenversion2)
	{
		while (i < lenversion1&&version1[i] != '.')
		{
			num1 = 10 * num1 + version1[i]-'0';
			i++;
		}

		while (j < lenversion2 && version2[j] != '.')
		{
			num2 = 10 * num2 + version2[j]-'0';
			j++;
		}

		if (num1>num2)return 1;
		else if (num1 < num2)return -1;
		i++;
		j++;
		num1 = 0;
		num2 = 0;
	}
	return 0;
}