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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
解法：
2，采用“.”分离时，要加上“\\.”
3，数组没有赋值时，默认为0
package Leetcode;
public class CompareVersionNumbers {
public int compareVersion(String version1, String version2) {
String[] ver1=version1.split("\\.");
String[] ver2=version2.split("\\.");
int maxlength;
if(ver1.length>=ver2.length){
maxlength=ver1.length;
}
else{
maxlength=ver2.length;
}
int[] ints1 = new int[maxlength];
for (int i=0; i < maxlength; i++) {
ints1[i] = Integer.parseInt(ver1[i]);
}
int[] ints2 = new int[maxlength];
for (int i=0; i < maxlength; i++) {
ints2[i] = Integer.parseInt(ver2[i]);
}
for(int i=0;i<maxlength;i++){
if(ints1[i]>ints2[i]){
return 1;
}
else if(ints1[i]<ints2[i]){
return -1;
}
}
return 0;
}
}
leetcode Compare Version Numbers
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原文地址：http://www.cnblogs.com/lilyfindjobs/p/4186486.html
