LeetCode: 165. Compare Version Numbers 比较版本号

骑猪看日落 2021-06-24 16:10 305阅读 0赞

**试题：**  
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.

Example 1:

Input: version1 = “0.1”, version2 = “1.1”  
Output: -1  
Example 2:

Input: version1 = “1.0.1”, version2 = “1”  
Output: 1  
Example 3:

Input: version1 = “7.5.2.4”, version2 = “7.5.3”  
Output: -1  
Example 4:

Input: version1 = “1.01”, version2 = “1.001”  
Output: 0  
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”  
Example 5:

Input: version1 = “1.0”, version2 = “1.0.0”  
Output: 0  
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to “0”

Note:

Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.  
Version strings do not start or end with dots, and they will not be two consecutive dots.  
**代码：**

class Solution {
        public int compareVersion(String version1, String version2) {
            String[] v1s = version1.split("\\.");
            String[] v2s = version2.split("\\.");
            int v1len = v1s.length, v2len = v2s.length;
            int maxlen = Math.max(v1len,v2len);
            
            for(int i=0; i<maxlen; i++){
                Integer v1 = i<v1len?Integer.valueOf(v1s[i]):0;
                Integer v2 = i<v2len?Integer.valueOf(v2s[i]):0;
                int com = v1.compareTo(v2);
                if(com!=0) return com;
            }
            return 0;
            
        }
    }