Compare Version Numbers

1 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 

.

 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

题目出处：https://leetcode.com/problems/compare-version-numbers/

2 解题思路

“.”将两个数字分割开，只需要比较相同位置上的两个数字的大小即可。如果相同，则比较下一个，直到不相同（或者没有）为止。有个特殊情况（“1.0”与“1”相等）。

3 相关知识

1.将数组转化为List的方法（将其作为参数传进去）为：

String[] v1s = version1.split("\\.");

ArrayList<String> alv1 = new ArrayList<String>(Arrays.asList(v1s));

2.字符串的split()方法的使用情况参照网址：http://www.cnblogs.com/mingforyou/archive/2013/09/03/3299569.html

4 源代码

package com.larry.easy;

import java.util.ArrayList;
import java.util.Arrays;

public class CompareVersionNumbers {
public int compareVersion(String version1, String version2) {
String[] v1s = version1.split("\\.");String[] v2s = version2.split("\\.");
ArrayList<String> alv1 = new ArrayList<String>(Arrays.asList(v1s));
ArrayList<String> alv2 = new ArrayList<String>(Arrays.asList(v2s));

//如果两个数组的个数不相同，较少的补零
int num = v1s.length - v2s.length;
if(num > 0){
for(int i = 0; i < num; i++) alv2.add("0");
}else if(num < 0){
for(int i = 0; num < 0; num++) alv1.add("0");
}

int size = alv1.size();
for(int i = 0; i < size; i++){
int first = Integer.parseInt(alv1.get(i));
int secd = Integer.parseInt(alv2.get(i));
if(first > secd) return 1;
else if(first < secd) return -1;
}
return 0;
}
public static void main(String[] args) {
String s1 = "0.1";
String s2 = "1.1";
String s3 = "0.1.0";
String s4 = "1";
String version1 = s4;
String version2 = s2;
System.out.println(new CompareVersionNumbers().compareVersion(version1, version2));
}
}