[LeetCode]Compare Version Numbers
发布日期:2021-11-22 02:48:53 浏览次数:2 分类:技术文章

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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  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;    }}

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