Leetcode 165 Compare Version Numbers

Compare numbers version1 and Version2.
If Version1 > Version2 return 1, if Version1 < Version2 return-1, otherwise return 0.

Assume that the version strings is non-empty and contain only digits and the. Character.
The. Character does not represent a decimal point and was used to separate number sequences.
For instance, 2.5 are not "both and a half" or "half to version three", it is the fifth Second-level revision of the SEC Ond first-level revision.

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

Solution:
The main idea was very simple and the code consists of three phases:
1.When Version1 and Version2 is 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; > 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(stringVersion1,stringVersion2) {inti =0;intj =0;intN1 = Version1.size ();intN2 = Version2.size ();intNUM1 =0;intnum2 =0; while(I < n1 | | J < n2) { while(I<n1 && version1[i]!='. ') {NUM1 = num1*Ten+ (version1[i]-' 0 ');            i++; } while(J<n2 && version2[j]!='. ') {num2 = num2*Ten+ (version2[j]-' 0 ');            j + +; }if(NUM1&GT;NUM2)return 1;Else if(NUM1&LT;NUM2)return-1; NUM1 =0; num2 =0;            i++;        j + +; }return 0; }};

Python solution:

 class solution:    # @param A, a string    # @param B, a string    # @return A Boolean     def compareversion(self, version1, version2):V1 = Version1.split ('. ') v2 = Version2.split ('. ') forIinchRange (max (len (v1), Len (v2))): Gap = (int (v1[i])ifI < Len (v1)Else 0)-(int (v2[i])ifI < Len (v2)Else 0)ifGap! =0:return 1 ifGap >0 Else-1        return 0

Leetcode 165 Compare Version Numbers