[LeetCode]Compare Version Numbers

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

Credits:

Special thanks to @ts for adding this problem and creating all test cases.

code:

public class Solution {
    public int compareVersion(String version1, String version2) {
    String[] s1 = version1.split("\\.");
		String[] s2 = version2.split("\\.");

		int i;
		for (i = 0; i < s1.length && i < s2.length; i++) {

			if (Integer.parseInt(s1[i]) == Integer.parseInt(s2[i])){
				
				continue;
			}
			if (Integer.parseInt(s1[i]) != Integer.parseInt(s2[i])){
				return Integer.parseInt(s1[i]) > Integer.parseInt(s2[i]) ? 1: -1;
			}
		}
	
		for (; i < s1.length; i++) {
			if (Integer.parseInt(s1[i]) != 0){
				return 1;
			}
		}
		for (; i < s2.length; i++) {
			if (Integer.parseInt(s2[i]) != 0)
				return -1;
		}
		return 0;
    }
}