• leetcode 165 Compare Version Numbers


    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.

    Here is an example of version numbers ordering:
    0.1 < 1.1 < 1.2 < 13.37

    解决方案:
    The main idea is very simple and the code consists of three phases:
    1.When version1 and version2 are not finished, compare the value of corresponding string before dot.
    2.If version1 is finished, check whether remaining version2 contains string not equal to 0
    3.If version2 is finished, check whether remaining version1 contains string not equal to 0

    Example1: version1==”11.22.33”, version2==”11.22.22”. 11 == 11; 22 == 22; 33 > 22; return 1.

    Example2: version1==”11.22.33”, version2==”11.22.33”. 11 == 11; 22 == 22; 33 == 33; return 0.

    Example3: version1==”11.22.33”, version2==”11.22.33.00.00”. 11 == 11; 22 == 22; 33 == 33; remaining version2 equals to 0; return 0.

    Example4: version1==”11.22.33.00.01”, version2==”11.22.33”. 11 == 11; 22 == 22; 33 == 33; remaining version1 contains 01; return 1.

    这里写图片描述

    class Solution {
    public:
        int compareVersion(string version1, string version2)
        {
            int i = 0;
            int j = 0;
            int n1 = version1.size();
            int n2 = version2.size();
    
            int num1 = 0;
            int num2 = 0;
    
            while(i < n1 || j < n2)
            { 
                while(i<n1 && version1[i]!='.')
                {
                    num1 = num1*10 + (version1[i]-'0');
                    i++;
                }
    
                while(j<n2 && 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;
    
        }
    };

    python解决方案:

    class Solution:
        # @param a, a string
        # @param b, a string
        # @return a boolean
        def compareVersion(self, version1, version2):
            v1 = version1.split('.')
            v2 = version2.split('.')
            for i in range(max(len(v1), len(v2))):
                gap = (int(v1[i]) if i < len(v1) else 0) - (int(v2[i]) if i < len(v2) else 0)
                if gap != 0:
                    return 1 if gap > 0 else -1
            return 0
    
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  • 原文地址:https://www.cnblogs.com/wangyaning/p/7853979.html
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