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leetcode------Compare Version Numbers
标题: | Compare Version Numbers |
通过率: | 14.8% |
难度: | 简单 |
Compare two version numbers version1 and version1.
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
这个题我看了好久没有看明白,我感觉这个题目没有描述清楚,没有说清楚版本号有可能是多级的比如说:1.1.1.1或者是4.0.1.1这样的给的示例让我认为都是
X.X这种的。我去看了别人的代码我才知道什么意思。。估计是我理解问题能力的吧,
对于两个字符串先用"."进行分片操作,分片完就有长度可以去判断了。用两个字符串长度长的值作为循环条件,(刚开始我用短的去处理,如果出现1.1 和1的问题还要去处理下面一个值,过于麻烦,直接用长的判断,如果越界就用0去补)进行诸位进行比较如果出现不同立即返回,如果一直的相同在循环条件结束后返回0即可。代码如下:
1 public class Solution { 2 public int compareVersion(String version1, String version2) { 3 if(version1==null || version2==null)return 0; 4 String [] str1=version1.split("\\."); 5 String [] str2=version2.split("\\."); 6 int len1=str1.length; 7 int len2=str2.length; 8 int len=len1>len2 ? len1:len2; 9 for(int i=0;i<len;i++){10 int num1= i<len1 ?Integer.parseInt(str1[i]):0;11 int num2= i<len2 ?Integer.parseInt(str2[i]):0;12 if(num1>num2)return 1;13 if(num1<num2)return -1;14 }15 return 0;16 17 }18 }
leetcode------Compare Version Numbers