[leetcode] Compare Version Numbers

问题描述：

Compare two version numbers version1 and
version1.

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

基本思想：

从高位到低位递归解决。 注意特殊case （1,1.0）;

代码：

public int compareVersion(String version1, String version2) { //java
if(version1.equals(version2))
return 0;

int fversion1 , fversion2;
String sversion1,sversion2;
if(version1.contains(".")){
int pos = version1.indexOf(".");
fversion1 = Integer.valueOf(version1.substring(0,pos));
sversion1 = version1.substring(pos+1,version1.length());
}
else {
fversion1 = Integer.valueOf(version1);
sversion1 = "0";
}

if(version2.contains(".")){
int pos = version2.indexOf(".");
fversion2 = Integer.valueOf(version2.substring(0,pos));
sversion2 = version2.substring(pos+1,version2.length());
}
else {
fversion2 = Integer.valueOf(version2);
sversion2 = "0";
}

if(fversion1 > fversion2)
return 1;
else if(fversion1 < fversion2)
return -1;
else return compareVersion(sversion1, sversion2);
}