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.
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：
:v1 = version1.split(‘.’)v2 = version2.split(‘.’)for i in range(max(len(v1), len(v2))):gap = (int(v1[i]) )if gap != 0:gap >
，也站在未路让我牵挂的人。
