LeetCode_Compare Version Numbers

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

题目属于简单级别，还是挺复杂的，思路：将字符串按照"."分割，将“.”之间的数字转化为整数，然后分别对应比较。例如：1.2和13.37，分别转化为数组[1,2]和[13,37]，从前往后对应比较1比13小，这时就可以得出结论1.2<13.37，若相等，则继续向后比较。这里需要注意，若两个字符串转化的数组不等长，需要将短的数组后面补0，例如，1.1和1.1.0，前者转化为[1,1]，后者是[1,1,0]，这时需要将前者补齐为[1,1,0]，然后再对应比较。代码如下：

public int compareVersion(String version1, String version2) {

		//.的个数

		int dot_num1 = 0;

		int dot_num2 = 0;

		for(int i = 0;i<version1.length();i++)

		{

			if(version1.charAt(i)=='.')

				dot_num1 ++;

		}

		for(int i = 0;i<version2.length();i++)

		{

			if(version2.charAt(i)=='.')

				dot_num2 ++;

		}

		//按照.的位置截取子串

		String arr1[] = new String[dot_num1+1];

		String arr2[] = new String[dot_num2+1];

		//记录每个字符串中的.的个数

		int index1[] = new int[dot_num1+2];

		int index2[] = new int[dot_num2+2];

		int in1 = 1;

		index1[0] = -1;

		for(int i = 0;i<version1.length();++i)

		{

			if(version1.charAt(i)=='.')

				index1[in1++] = i;

		}

		index1[in1] = version1.length();

		int in2 = 1;

		index2[0] = -1;

		for(int i = 0;i<version2.length();++i)

		{

			if(version2.charAt(i)=='.')

				index2[in2++] = i;

		}

		index2[in2] = version2.length();

		//截取子串

		int index = 0;

		for(int i = 0;i<index1.length-1;i++)

		{

		  	arr1[index++] = version1.substring(index1[i]+1,index1[i+1]);

		}

		index = 0;

		for(int i = 0;i<index2.length-1;i++)

		{

			arr2[index++] = version2.substring(index2[i]+1,index2[i+1]);

		}

		int len1 = arr1.length;

		int len2 = arr2.length;

		int max = len1>len2?len1:len2;

		//所有子串转化为整数形式

		int num1[] = new int[max];

		int num2[] = new int[max];

		for(int i = 0;i<len1;i++)

			num1[i] = Integer.parseInt(arr1[i]);

		for(int i = 0;i<len2;i++)

			num2[i] = Integer.parseInt(arr2[i]);

		//长度不一，短的后面0补齐

		/*

		 * *

		 * 例如1.0和1.0.0

		 * 前者生成的num1数组为[1,0]

		 * 后者生成的num2数组为[1,0,0]

		 * 这里将num1补齐为[1,0,0],便于比较

		 */

		if(len1>=len2)

		{

			for(int i = len2;i<len1;i++)

			{

				num2[i] = 0;

			}

		}

		else

		{

			for(int i = len1;i<len2;i++)

			{

				num1[i] = 0;

			}

		}

		/*for(int i = 0;i<num2.length;i++)

		{

			System.out.println(num2[i]);

		}

		for(int i = 0;i<num1.length;i++)

		{

			System.out.println(num1[i]);

		}*/

		for(int i = 0;i<max;i++)

		{

			if(num1[i]>num2[i])

				return 1;

			if(num1[i]<num2[i])

				return -1;

		}

		return 0;

	}

巴特西

LeetCode_Compare Version Numbers

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