标签:Compare return version1 version2 number Version 165 revision String
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.
You may assume the default revision number for each level of a version number to be 0
. For example, version number 3.4
has a revision number of 3
and 4
for its first and second level revision number. Its third and fourth level revision number are both 0
.
题目大意:
版本号的比较
解法:
使用字符串分割得到两个字符串数组。再按每一位字符串数组的整型进行比较。
java:
class Solution { public int compareVersion(String version1, String version2) { String []str1=version1.split("\\."); String []str2=version2.split("\\."); int i=0; while(i<Math.min(str1.length,str2.length)){ if(Integer.parseInt(str1[i])<Integer.parseInt(str2[i])) return -1; else if(Integer.parseInt(str1[i])>Integer.parseInt(str2[i])) return 1; i++; } while(i<str1.length){ if(Integer.parseInt(str1[i])>0) return 1; i++; } while(i<str2.length){ if(Integer.parseInt(str2[i])>0) return -1; i++; } return 0; } }
代码可以简化为:
class Solution { public int compareVersion(String version1, String version2) { String []str1=version1.split("\\."); String []str2=version2.split("\\."); int length=Math.max(str1.length,str2.length); for(int i=0;i<length;i++){ Integer v1=i<str1.length?Integer.parseInt(str1[i]):0; Integer v2=i<str2.length?Integer.parseInt(str2[i]):0; int compare=v1.compareTo(v2); if(compare!=0) return compare; } return 0; } }
标签:Compare,return,version1,version2,number,Version,165,revision,String 来源: https://www.cnblogs.com/xiaobaituyun/p/10758467.html
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