其他分享
首页 > 其他分享> > LeetCode-165-Compare Version Numbers

LeetCode-165-Compare Version Numbers

作者:互联网

算法描述:

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.

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:

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

解题思路:两个指针同时向后遍历,将两个'.' 之间的所有字符转化成数字,然后进行比较。

    int compareVersion(string version1, string version2) {
        int l1 = version1.size();
        int l2 = version2.size();
        int i=0;
        int j=0;
        int num1=0;
        int num2=0;
        
        while(i < l1 || j < l2){
            while(i < l1 && version1[i]!='.'){
                num1 += num1*10 + (version1[i]-'0'); 
                i++;
            }
            
            while(j < l2 && version2[j]!='.'){
                num2 += num2*10 + (version2[j]-'0');
                j++;
            }
            
            if(num1 > num2) return 1;
            else if(num1 < num2) return -1;
            
            num1=0;
            num2=0;
            i++;
            j++;
        }
        return 0;
    }

 

标签:Compare,version1,version2,level,int,number,Version,165,revision
来源: https://www.cnblogs.com/nobodywang/p/10399648.html