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;
}
}
leetcode [165]Compare Version Numbers
原文：https://www.cnblogs.com/xiaobaituyun/p/10758467.html