Compare Version Numbers leetcode

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

这道题的trick在于 version 1.1.20 和1.1.2 是相同的.

这里用到String.split() 来按照"." 把string 分成array 

String.split() accepts a regular expression (regex for short) and dot is a special char in regexes. It means "match all chars except newlines". So you must escape it with a leading backslash. But the leading backslash is a special character in java string literals. It denotes an escape sequence. So it must be escaped too, with another leading backslash. Like this:

fileName.split("\\.");

public class Solution {
    public int compareVersion(String version1, String version2) {
        String[] arr1 = version1.split("\\.");
        String[] arr2 = version2.split("\\.");
        int i = 0;
        for (; i < arr1.length && i< arr2.length; i++) {
            if (Integer.parseInt(arr1[i]) < Integer.parseInt(arr2[i])) {
                return -1;
            } else if (Integer.parseInt(arr1[i]) > Integer.parseInt(arr2[i])) {
                return 1;
            }
        }
        while (i < arr1.length) {
            if (Integer.parseInt(arr1[i++]) > 0) {
                return 1;
            }
        }
        while (i < arr2.length) {
            if (Integer.parseInt(arr2[i++]) > 0) {
                return -1;
            }
        }
     
        return 0;
    }
}