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

思路：
首先，以"."为分隔符进行字符串拆分，但要注意，在Pattern类中“.”表示任意字符，而"\\."用来表示小数点；

然后，逐个比较拆分后字符串的大小（转换为Int型），因为对于“01”和“1"被认为是相等的。注意，还要考虑拆分成的字符数组不等长的情况，如

"1"<“1.1”；但“1”和“1.0”却是相等的。

public int compareVersion(String version1, String version2) {
String[] v1=version1.split("\\.");
String[] v2=version2.split("\\.");//在Pattern类中“.”表示任意字符，而"\\."来表示小数点
System.out.println(v1.length);
System.out.println(v2.length);
int N=Math.min(v1.length,v2.length);
int i=0;
for( i=0;i<N;i++){
if(Integer.parseInt(v1[i])<Integer.parseInt(v2[i])) return -1;
else if(Integer.parseInt(v1[i])>Integer.parseInt(v2[i])) return 1;
else continue;
}
if((v1.length<v2.length)&&Integer.parseInt(v2[i])!=0) return -1;//注意考虑另个字符串拆分之后长度不等的情况，但注意1.0与1是相等的
if((v1.length>v2.length)&&Integer.parseInt(v1[i])!=0) return 1;
else return 0;
}