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

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解题思路：

常规思路，用split()分割string, IMPORTANT : Some special characters need to be escaped while providing them a delimiters like "." and "\".

然后用Integer.parseInt(x)转换成int 比较大小。注意： 1.0 和1 相等，因此判断还剩一个string 的时候，值要不等于0 才大于另一个string.

题目不难，但容易错，要小心仔细。

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 Java code:

public int compareVersion(String version1, String version2) {
        String delimiter = "\\.";
        String[] v1 = version1.split(delimiter);
        String[] v2 = version2.split(delimiter);
        int len = Math.max(v1.length, v2.length);
        for(int i = 0; i< len; i++){
            if(i < v1.length && i < v2.length){
                if(Integer.parseInt(v1[i]) > Integer.parseInt(v2[i])) {
                    return 1;
                }else if(Integer.parseInt(v1[i]) < Integer.parseInt(v2[i])){
                    return -1;
                }
            }else if(i < v1.length) {
                if(Integer.parseInt(v1[i]) > 0) {
                    return 1;
                }
            }else if(i < v2.length){
                if(Integer.parseInt(v2[i]) > 0) {
                    return -1;
                }
            }
        }
        return 0;
    }

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Reference:

1. http://www.programcreek.com/2014/03/leetcode-compare-version-numbers-java/