leetcode 152: Compare Version Numbers

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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

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

public class Solution {
    public int compareVersion(String version1, String version2) {
        if(version1 == null || version2 ==null) return 0;
        
        String[] v1 = version1.split("\\.");
        String[] v2 = version2.split("\\.");
        
        int n1 = v1.length;
        int n2 = v2.length;
        
        int i=0;
        
        while(i<n1 || i<n2) {
            int x1 = i<n1 ? Integer.parseInt(v1[i]) : 0;
            int x2 = i<n2 ? Integer.parseInt(v2[i]) : 0;
            if(x1 > x2) return 1;
            else if(x1 < x2) return -1;
            else ++i;
        }
        
        return 0;
        
     }
}

leetcode 152: Compare Version Numbers

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原文地址：http://blog.csdn.net/xudli/article/details/42081113