Here is an example of version numbers ordering: 0.1 < 1.1 < 1.2 < 13.37 比较两个字符串数字的大小:(小数点可能不止一个) 注意的地方: 1、多个小数点的处理。 2、考虑清楚细节就可以了,题目本身不难。 1、直接判断,代码较多。 publicclassSolution {publicintcompareVersion(String version1, String versio...
步骤2: 从字符串中提取数字 接下来,我们需要从字符串中提取数字。可以使用Java的正则表达式来实现这一步骤。以下是具体的代码: importjava.util.Scanner;publicclassCompareNumbers{publicstaticvoidmain(String[]args){Scannerscanner=newScanner(System.in);System.out.println("请输入第一个字符串:");Stringstr1=sca...
In Java programming, the comparison of integers is a fundamental aspect, serving as the basis for decision-making and logical operations. One common approach is to use relational operators, such as<,<=,>,>=,==, and!=, to establish relationships between numbers. ...
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 解题思路: 先进行split,然后转为int比较即可,JAVA实现如下: 1 2...
Compare Version Numbers LC解题记录 题目内容 比较不同的版本号,并根据大小返回-1,1或0。并提醒2.5版本意思是第二代的第五次升级,反正不是数字上的2.5的意思。 解决思路 直观的想法是,找到比较两个字符串(版本号)的方法,举个例子,1.1.2和1.2.3的比较,那么先比较第一位,两个1相等。那么就再看第二位,1<...
0 1 -1 0 Example 3: Here is a general example where the user can enter the numbers of his choice and get the desired output. Thetryandcatchblock is used for handling any exception taking place during the execution of the program. (For example, the user enters a String, etc. in place...
Float.compare(float f1, float f2)和Float.compare(double d1, double d2)的内部的逻辑处理基本一致。具体步骤:先比较他们的大小;如果,值不是简单的大于小于关系的话,需要转为类型在进行比较;一般情况是0.0、-0.0这种特殊的情况。最后运用两个三元表达式进行值的比较; ...
System.out.println("f1 and f2 are not equal using BigDecimal\n"); } Program Output. f1 =1.1 f2 =1.1 f1 and f2 are equal using BigDecimal That’s all aboutcomparing floating point numbers in java. Share your thoughts in comments section. Happy Learning !!
(v2[i]) > 0: return -1 return 0 Reference https://leetcode.com/problems/compare-version-numbers
Using subtraction is a very common idiom for comparing numbers in sorting routines, and is almost certainly the reason why compareTo() is defined to return an integer in the first place. Indeed, you may be wondering why we bothered with the previous version of doing comparisons and returning ...