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/*

Author: King, wangjingui@outlook.com

Date: Dec 15, 2014

Problem: Compare Version Numbers

Difficulty: Easy

Source: https://oj.leetcode.com/problems/compare-version-numbers/

Notes:

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

Solution: ...

*/

public class Solution {

public int compareVersion(String version1, String version2) {

long a = 0, b =0;

int v1len = version1.length(), v2len = version2.length();

int i = 0, j = 0;

while (i < v1len || j < v2len) {

a = 0; b =0;

while (i < v1len && version1.charAt(i) != '.') {

a = a * 10 + version1.charAt(i) - '0';

++i;

}

++i;

while (j < v2len && version2.charAt(j) != '.') {

b = b * 10 + version2.charAt(j) - '0';

++j;

}

++j;

if (a > b) return 1;

if (a < b) return -1;

}

return 0;

}

}