Java [Leetcode 165]Compare Version Numbers
2016-02-22 14:54
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题目描述:
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
The
For instance,
Here is an example of version numbers ordering:
解题思路:
以小数点分开,逐个比较。
代码如下:
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,
2.5is 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
解题思路:
以小数点分开,逐个比较。
代码如下:
public class Solution { public int compareVersion(String version1, String version2) { String[] levels1 = version1.split("\\."); String[] levels2 = version2.split("\\."); int length = Math.max(levels1.length, levels2.length); for(int i = 0; i < length; i++){ Integer v1 = i < levels1.length ? Integer.parseInt(levels1[i]) : 0; Integer v2 = i < levels2.length ? Integer.parseInt(levels2[i]) : 0; int compare = v1.compareTo(v2); if(compare != 0) return compare; } return 0; } }
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