Compare Version Numbers (Java)

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最新推荐文章于 2022-03-31 10:46:16 发布

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分类专栏： LeetCode easy String 文章标签： LeetCode

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本文详细阐述了如何通过编程实现版本号之间的比较，包括解析版本号字符串、转换为整数进行比较，并提供了实例代码演示。

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

比较版本号大小，把'.'分开的部分分别比较，每部分string转为十进制，注意有"1.1.1.1"这种多个'.'的情况

Source

	  public int compareVersion(String version1, String version2) {
	       int i = 0, j = 0;
	       
	       while(i < version1.length() || j < version2.length()){
	    	   int a = 0, b = 0;
	    	   while(i < version1.length() && version1.charAt(i) != '.'){   //判越界的要放在前面 否则放在后面容易报抛出异常 比如输入1,0时
	    		   a = a * 10 + (version1.charAt(i) - '0');    //注意减去'0'转换为数字
	    		   i++;
	    	   }

	    	   while(j < version2.length() && version2.charAt(j) != '.'){
	    		   b = b * 10 + (version2.charAt(j) - '0');
	    		   j++;
	    	   }
	 
	    	   i++;
	    	   j++;
	    	  
	    	   if(a > b) return 1;
	    	   if(a < b) return -1;
	       }
	       return 0;
	    }

Test

	  public static void main(String[] args){
		  String a = "01";
		  String b = "1";
		  System.out.println(new Solution().compareVersion(a, b));
	  }